Perhaps it would be interesting to dissect the circuit a bit, to see why it can't work (especially the right side) as well as a little analysis of how/why it does work as (for example) Tony Stewart has drawn it.
So let's start by looking only at the left side:
simulate this circuit – Schematic created using CircuitLab
Given an appropriate value of R1, this much can work. Current flows through R1 to the base of Q3, turning it on and allowing current to flow through the LED to ground, so the LED lights up.
When the switch closes, the current flowing through R1 is now conducted directly to ground, so no current flows into the base of Q1. Q1 turns off, no current flows through the LED. So far so good.
The right half of your circuit has a much bigger problem though. R2 is drawn as a 0K resistor, so the right half of your circuit really looks like this:
simulate this circuit
Regardless of what happens on the left side, the base of Q1 is tied to ground. No current flows in, and it never turns on. Regardless of the switch on the left, no current flows into the base of Q1, and the LED never turns on.
With the circuit as Tony Stewart re-drew it, the situation changes completely. In this case, when the switch is closed, Q3 is turned off, so there's a high voltage at the collector. Current can flow through the LED, through R1, and into the base of Q1. So, when Q3 is turned off, Q1 turns on. R1 has to be selected to be small enough that Q1 turns on, but large enough that we're not drawing enough current for the LEDs on the left to light up (at least enough to notice). Since Q1 (presumably) has a fair amount of gain, that's usually going to be a pretty broad range though (you don't need much current flowing into the base to allow a fair amount of current to flow between the emitter and collector).
If we open the switch, Q3 turns on, so the left side of R1 is pulled to ground. Current stops flowing into the base of Q1, so the LEDs on the right turn off.