The four players in a game form 6 pairs. Since any pair plays in at least one game, the number of all pairs $\binom{n}{2} = \frac{n(n-1)}{2}$ 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.