Pickup basketball teams

The number of ballers from each of Blacksburg, Greensboro and Silver Spring who show up to a pickup basketball game each week is identically and independently distributed as the uniform distribution on $\{1, 2, 3, 4, \dots, N\},$ for some integer $N.$

What is the probability that, on any given week, it’s possible to form two equal teams with everyone playing, where two towns are pitted against the third?

Let $B,$ $G$ and $S$ be the number of players who showed up from Blacksburg, Greensboro and Silver Spring, respectively. Without loss of generality, let's assume that $S$ is the largest and we seek the number of configurations where $B+G=S.$

First note that if $S$ is the maximum, then no such configurations can occur if $S=1(=B=G).$ For any $S = 2, \dots, N,$ there are $S-1$ different combinations of integers $B$ and $G,$ such that $B+G=S;$ those being $(B,G) \in \{ (1,S-2), (2, S-3), \dots, (S-3, 2), (S-1, 1) \}.$ So the total number of games where $B+G=S$ is $$\sum_{S=2}^N (S-1) = \frac{N(N-1)}{2}.$$
From symmetry, we get three times this number (since we arbitrarily assumed that $S$ was the maximum, but it could equally likely be $B$ or $G$), so the total number of configurations where two equal teams where two towns are pitted against the third are $3 \frac{N(N-1)}{2}.$ The total number of possible configurations of players to show up on a given week is $N^3,$ so the probability is $$p_N = \frac{3(N-1)}{2N^2}.$$
In particular, if $N = 5,$ then the probability is $p_5 = 24\%.$