I joined this forum yesterday, after I came across your interesting discussion in Google.
Your thoughts are wonderful and I fully support them. My point is just that they are based more on a detailed and sometimes formal analysis of the INIC circuit (what it does) than on the disclosure of its philosophy (why it does this). So I will try to roughly fill that gap with my comment.
We can consider this circuit from two perspectives: first - as a circuit with only input and no output (a load with negative resistance); second - as a circuit with input and output (an amplifier with mixed feedback).
Negative load. Beginning from the early 90's, I spent a lot of effort to reveal and explain in an easy and intuitive way the first perspective. If you are interested and patient enough, you can familiarize yourself with the resources I created in Web; I described them in detail in two questions asked by me in ResearchGate - What is negative impedance? and What is the basic idea behind the negative impedance converter? For those who do not have patience to read all of this, here is a very brief explanation.
The circuit behaves as an active load (dynamic voltage source with internal resistance R) that reverses the current through the resistor R (in the original Wikipedia picture) and "pushes" it back to the input source. In this way, it converts the resistor R (originally consuming a current) into a negative "resistor" -R (producing a current). It does this by opposing (through the resistor) a reverse and higher (2V) voltage to the input voltage (V).
This is the output voltage of the operational amplifier and it is not used here... but still the circuit has an output... and, although it sounds strange, it is its input! Simply the circuit behaves like a source that attacks back the input source...
Amplifier with mixed feedback. According to me, this is the subject of the question asked here. As described in the comments above, this circuit is an amplifier with negative feedback, which is partially neutralized by a weaker positive feedback. But what is the point of that?
In general, the positive feedback increases the gain of the imperfect amplifiers and it is used in the past (remember Armstrong's regenerative idea). But in our case, the op-amp has a huge gain and this is not necessary. Then what is the point of using positive feedback here?
My speculation is that we can use it to decrease the ratio R3/R4 (in the second figure) in the case of INIC or R2/R1, in the case of VNIC (when the input voltage is applied to the inverting input). As a result, the resistors R2 and R3 can be low resistive.
In this amp application, the op-amp output is the circuit output. But as above, this amplifier has another output... and this is its input... so the circuit can act as an exotic 1-port amplifier...
Reinventing INIC by CircuitLab experiments
I decided to expand on my answer from nine years ago by revealing the secret of the "Wikipedia circuit" because it is not just a circuit, but rather a concept. It represents one of the possible ways to obtain a negative resistance from a positive one by reversing the current direction; hence the name current-inversion negative impedance converter (INIC). Legendary circuits and techniques such as Howland current source, load canceller, bootstrapping, etc. are built on it. So let's see what this powerful idea is with the help of which we can modify resistance.
I suggest that we do it in the form of step-by-step CircuitLab experiments with which we can follow the evolution of the idea.
Positive resistance
In the beginning, a 1 kΩ "positive" resistor consumes 1 mA current from 1 V voltage source.
simulate this circuit – Schematic created using CircuitLab
Increased resistance
The straightforward way to increase (eg, double) its resistance is to replace it with a 2 kΩ resistor. But we can apply a small trick to artificially increase the resistance connecting in series and in the opposite direction a "behavioral" voltage source V1 with 0.5*V voltage. It is subtracted from the main voltage V and the current decreases twice. So the main source V has the illusion that the resistance has doubled (R' = V/I = 2 kΩ).
simulate this circuit
A more practical implementation is through an "amplifier" with a gain of 0.5 (I have set such a gain to an op-amp from the CircuitLab library).
simulate this circuit
Infinite resistance
If we continue to increase the additional voltage V1, the current will decrease more and more and the resistance R' will increase. When the two voltages equalize, the current stops flowing and the resistance R' becomes infinite (we can simulate it by a following behavioral voltage source V1 = V).
simulate this circuit
This famous circuit trick figuratively named "bootstrapping", can be implemented by an amplifier with a fixed gain of 1...
simulate this circuit
... that is usually made by an op-amp with a 100% negative feedback (op-amp follower).
simulate this circuit
Negative resistance
And now comes the most interesting thing, for the sake of which we did everything up to here. If we continue to increase the additional voltage, e.g. we double it (V1 = 2*V), the current reverses its direction and enters the source V as if the resistance does not consume but produces current... it has become negative (R' = V/-I = -1 kΩ).
simulate this circuit
We can implement it by an amplifier with a fixed gain of 2...
simulate this circuit
... that is made by an op-amp non-inverting amplifier.
simulate this circuit
Is there feedback?
Neither positive nor negative feedback
The amplifier output is connected through the resistor R to its non-inverting input, but this does not automatically mean that there is positive feedback. If the input voltage source is perfect (with zero internal resistance), the amplifier output will not be able to change the voltage at the non-inverting input, and there will be no positive feedback. Since the amplifier is with fixed gain, there is no need for negative feedback.
Only positive feedback
If the input voltage source is imperfect (with some internal resistance) or there is an input resistor in series, then the amp output can change the non-inverting input voltage; so there is positive feedback. For stability, the feedback loop gain must be less than 1. As above, there is no need for negative feedback. See for example Schematic 4.2.
Both positive and negative feedback
The op-amp non-inverting amplifier implementation needs negative feedback. For stability, it must dominate over the positive feedback. See for example Schematic 4.3 and the application below.
Applications
Howland current source
Now that we have uncovered the secret of INIC, let's consider one of its most famous applications - the Howland current source. Let's follow the evolution of this great idea.
"Perfect" current source: The simplest current source is just a resistor (Rin) in series to a voltage source (Vin). If we measure its current by a perfect ammeter (with zero resistance), the current is exactly as Ohm's law says - IL = Vin/Rin = 1 mA.
simulate this circuit
Imperfect current source: But if we measure the current by an imperfect ammeter (e. g. with 1 kΩ resistance RL), the current will be lower - IL = Vin/(Rin+RL) = 0.5 mA.
simulate this circuit
Let's sweep RL from zero to 10 kΩ to see its impact on the current.
Improved current source: The Howland idea is brilliant: to neutralize the 1 kΩ positive resistance (Rin) with -1 kΩ negative resistance (R1, R2, R and OA). The result is infinite resistance, i.e. constant load current IL. Let's investigate it by sweeping the load (ammeter) resistance RL from 0 to 10 kΩ. Note that in the schematic below, RL = 1 kΩ (to see or change it, open the IL parameters window).
simulate this circuit
As you can see in the graph below, when RL increases from 0 to 10 kΩ, the voltage across the load increases from 0 to 10 V but the current does not change. Do not pay attention to the yellow curve; it is only to "cheat" the simulator autoscaling.
Helped current source: The trick behind the Howland current source is really clever - making a perfect current source (the whole circuit) by helping an imperfect current source (Vin and Rin) with another auxiliary current source (R1, R2, R and OA).
In the graph below, when the load resistance increases, the input current decreases but the "helping" current increases. As a result, their sum - the load current, stays constant.
Load canceller
Here is an even more extravagant INIC application where a positive resistance is neutralized by an equivalent negative resistance.
Unloaded voltage divider: The humble potentiometer works precisely if it is not loaded (the voltmeter is perfect with extremely high resistance).
simulate this circuit
If the wiper is in the middle (K = 0.5), Vout = 0.5Vin.
Loaded voltage divider: But if we connect a low-resistance (only 100 Ω) load RL, the current increases and the output voltage drastically decreases...
simulate this circuit
... when Vin changes from zero to 10 V.
Helped voltage divider: The remedy is the same as in the Howland current source - to help the weak voltage divider by supplying the load with another current source. In terms of resistance, the 100 Ω positive resistance RL is completely neutralized by the 100 Ω negative resistance of the INIC (R1, R2, R and OA).
simulate this circuit
As a result, as though there is no load connected to the voltage divider, and the graph is the same as in the Schematic 5.2.1.