The two transistors make a latch circuit. When the base of Q1 rises above Vbe (about 0.6V) it's collector voltage drops and the relay operates. At the same time, Q2 is brought into conduction and provides more base current to Q1 so it stays in 'on' condition. Without the capacitor, the latch would react to any rise in voltage above Vbe and trigger the circuit, even it if was only for a few uS length. What I am saying is the state of the comparator will be indeterminate as the supply voltage rises when the power is turned on and might trigger the latch. The capacitor makes it less responsive so a longer 'high' from the comparator is needed before the latch operates.
What isn't clear from the schematic is what the switch connected to the LED does. If it is a second set of contacts on the relay it stops Q1 and Q2 being a latch and makes them a monostable. When a long enough high comes out of the comparator, the capacitor charges enough to make Q2 conduct, it operates the relay and that disconnects the comparator so no further base current can be provided. There would then be a short delay as the capacitor discharged into Q1 before the relay turned off again. If the comparator output has gone low, it will await a further high before repeating the process, if the comparator output is still high, the circuit will oscillate with the relay turning on and off cyclically.
Brian.