This is actually a useful circuit- for low frequencies it acts as a precision rectifier.
For inputs < 0V, the output is -0.5*Vin
For inputs > 0V, the output is +0.5*Vin (if loaded with a 15K resistor to ground or a virtual ground, which I suspect your inverting stage has-- 15K and 30K to yield an output voltage of -|Vin|).
You always get a positive output voltage from this circuit fragment. The 15K you mention is just to allow the op-amp to swing very close to the negative rail (ground).
If the input voltage is less than zero, it acts as an inverting amplifier, the op-amp output drives the diode anode to a voltage equal to -0.5*Vin (the op-amp output itself will be a bit higher to account for the diode drop).
If the input voltage is greater than zero, the op-amp saturates at the negative rail (ground) the diode is reverse biased, and the circuit looks like a 15K resistor. Hence, if you load it with 15K it will have an output of 0.5*Vin (voltage divider).
In the below top schematic, the shown (different) Vout = -|Vin|. Of course the second op-amp requires a negative supply, but the first one does not.
In the below bottom schematic, Vout = +|Vin| and neither op-amp requires a negative supply.
simulate this circuit – Schematic created using CircuitLab