The problem statement:
Suppose $n$ players engage in a tournament in which each player plays every other player in exactly one game, to a win or a loss. Let $w_i$ and $l_i$ denote the wins and losses of the $i$th competitor, $i = 1, 2, ... n$. Prove that $\sum {w_i}^2 = \sum {l_i}^2$.
A rather un-illuminating proof of this is the following (WLOG we treat $n= 4$):
Each player has exactly $4-1 = 3$ wins and losses total, so $w_i + l_i = 3$, and there will be an equal number of wins and losses total as well (in fact, $4 \choose 2$). Thus
$${w_1}^2 - {l_1}^2 + {w_2}^2 - {l_2}^2 + {w_3}^2 - {l_3}^2 + {w_4}^2 - {l_4}^2 = 0$$ $$ \iff 3 (w_1 - l_1 + w_2 - l_2 + w_3 - l_3 + w_4 - l_4) = 0$$
which is true.
I wanted to ask: is there a combinatorial proof of this which explains it better? Something with pigeonholing? (This might be pointless, but I still don't "get" this problem. Perhaps there's nothing else to "get".)