Wiring a 2.5km football field with 220v
It may be simpler than you think. 

The wall adapters you're using are 24V@0.5A, which is 12W. Assuming 0.8 efficiency, that's 15W that mains has to provide. Under 220V, that's I=68.2mA*. 

You have 15 adapters distributed along 2.5km, that's 166.7m of wires between each adapter. AWG10 wires have a resistance of 3.277Ohms/km, so on each line there will be R=546.3mOhms between each adapter. As shown on the equivalent circuit below (where I've modelled 3 adapters out of 15), the voltage drop experienced by each adapter will be twice the voltage drop in the line resistances (single and return).







simulate this circuit &ndash; Schematic created using CircuitLab

Note that the resistance values for the adaptors are for illustration purposes only (they're likely to change in reality), worst case current 'I' calculated above will be used instead. This is simplified this way:







simulate this circuit

In your case, there will be 15 resistors. Let i be the index of the adapter considered, in your case between 1 and N=15. The total voltage drop at adapter i is therefore:

$$\Delta V(i)=\sum\limits_{k=0}^{i-1} 2R(N-k)I$$
$$i \in 1,2,...,15$$

The 15th adapter will have the biggest drop, but it is only:


Or 4% of the input voltage. Since your adapters work from 100 to 220V, you're fine. No need to do anything other than wiring them around the football field. 

Just to check, the maximum power dissipated (~in terms of density) is in the first segment before the first adapter:
$$P_{dissipated}=\frac{\Delta V(1)^2}{2*R}=1W$$
Or 0.6mW every 10cm - not a heater, not a problem.

You should probably check the efficiency of the wall adapters and redo the calculations but it shouldn't be so bad that all this is thrown out the window.

*: Only if the voltage drop is neglectable compared to 220V. Otherwise, it's an iterative process: a second iteration with I=Power_device/(220-total voltage drop)=71.1mA will give you a more accurate voltage drop value, then a third one an even more accurate etc.
In this case I got 1) 8.94V 2) 9.32V 3) 9.335V; I think it converges to about 9.34V. This will actually be an overestimate since it is assumed all taps draw the same current (the biggest, after the total voltage drop) when in fact they draw less and less as you get closer to the supply.