Spring Mathematical Tournament

The four players in a game form 6 pairs. Since any pair plays in at least one game, the number of all pairs $\left(\genfrac{}{}{0ex}{}{n}{2}\right)=\frac{n(n-1)}{2}$ do not exceed 6 times the number of the games, i.e., $6n$. Hence $\frac{n(n-1)}{2}\le 6n$ which is equivalent to $n\le 13$.

For $n=13$ let $1,2,\dots ,13$ be the numbers of the players. A possible distribution with 13 games is the following: players $i,i+2,i+3,i+7$ (modulo 13) play in game number $i$, $1\le i\le 13$. It is easy to see that any two players play in one and the same game exactly once.

Hence the maximal value of $n$ equals 13.