Quick calculations with no guarantees about accuracy:

1 Btu = energy that must be added or removed to change temperature of one pound of water by one degree Fahrenheit.

20,000 gallons of water = (approx) 1,700,000 lb.

At an arbitrarily chosen 90 F water temperature, and assuming that the surface water is at about the same temperature as the adjacent air, the partial pressure of water vapor evaporated from the surface by an efficient process will be approximately 1.5 in Hg and the "latent heat of fusion" taken from the water and absorbed into the vapor will be on the order of 444 Btu/lb, meaning that evaporation of 1 lb of water at 90 F will remove about 400 Btu from the water left in the pool. (Evaporation of one gallon of water might remove 34,000 Btu, if the evaporation is "efficient.")

Removal of the of heat carried off by evaporation of one pound of water will cool the 1,700,000 lb of water in the pool by 0.000235 degrees F.

To have 1,700,000 lb of water left in the pool all cooled by 1F would require evaporation of about 4,250 pounds of water, or approximately 500 gallons – assuming that there is no other heat exchange occuring.

If the pool seems uncomfortably warm, it is likely that it is being heated at the surface mainly by solar insolation. If you can find an average insolation rate (most likely watts/sq meter) and multiply by the surface area of the pool you should be able to estimate how much water gets boiled out of the pool just to maintain the equilibrium temperature that you see without additional cooling. You might want to add that water loss rate to what you'd add to evaporate enough to get the cooling you think you want. To keep your cooler water at the lower temperature, you would need to assure that you continue to evaporate enough water at a rate that compensates for the incoming solar heat, since at a lower temp the natural evaporation will be reduced.

In "artificial" pools, it's quite common for the top layers of the water to be significantly warmer than the overall average temperature of the pool, and since you swim mostly at the surface it feels like the whole pool is hot. The "layer" of significantly warmer water near the surface often is only one or two feet thick, with cooler water below it. It generally takes 15 to 30 ft depth to get the "max density 34F bottom water," but even at 6 ft you may see as much as 10F lower temp than at the surface - depending on what circulation you have installed.

A sufficiently clever person might be able to "survey" the actual temperatures at different depths to determine an "average pool water temperature" and if the average is much lower than the surface layer temp, simply circulating (bottom to top) would make using the pool more comfortable. Drawing water from the bottom of the deepest part of the pool, and spraying or just squirting it onto the top water is likely to be effective, and evaporation from the "squirter" (or a waterfall) may provide some additional effect. The other method is to just add more kids (and throw in a couple of dogs and a goat or two) to stir things up.

I can't see much prospect for actually gaining much by simulating a "swamp cooler," but a healthy spray/fountain to move the water around some is fairly likely to help, and you should get about as much evaporative cooling as with a fancier setup. For the water volume indicated, you might want something like a 30 gpm pump(?), but it's hard to make much of an estimate without knowing the geometry of the pool, and something much smaller might work as well.

Just don't get a pump so big it knocks mama on her a**.

John