ingtar33 :
3) using a robust custom water cooling solution, move the radiators for the water loop outside your room, perhaps outside the house, or into a better ventilated room. This solution would require some construction work and serious thought but people have done it successfully.
A friend was in a similar situation (no AC, no ventilation, had to run a computer) and we got to talking about possible solutions. One idea we came up with was to get a water cooler and just stick the radiator into a 5 gallon bucket of water.
Air at 20 C (68 F) has a density of 1.2 kg/m^3. If your room is 20x20x8 ft (90.6 m^3), it has 108.7 kg of air. Air has a specific heat of 1.005 kJ/kg*K. For 108.7 kg of air, that's 109.3 kJ/K.
If the room is unventilated and your computer is burning 500 Watts, in 1 hour the air temperature will rise (0.5 kJ/s)*(3600 s) / (109.3 kJ/K) = 16.47 K or almost 30 degrees F. So if the room air starts at 68 F, after an hour it'll be 98 F. (Water vapor in the air also absorbs some heat energy and decreases this temperature change, but your body gives off about 100 Watts and increases this temperature change.)
Liquid water has a specific heat of 4.186 kJ/kg*K. A 5 gallon bucket holds just under 40 lbs of water (18 kg). So it can absorb (4.186 kJ/kg*K)*(18 kg) = 75.348 kJ/K.
So if the computer is consuming 500 Watts, in 1 hour the water temperature will rise (0.5 kJ/s)*(3600 s) / (75.348 kJ/K) = 23.9 K = 43 degrees F. So if the water comes out of the tap at 20 C (68 F), after an hour the water in the bucket will be 44 C (111 F), and the air remains its original temp. After an hour, just empty the bucket into the toilet or bathtub and fill it with a new batch of cold water. Or if you want to game for multiple hours, just get multiple buckets of water and either move the radiator from one bucket to the next, or get a fishtank pump and circulate the water between buckets.