Solution:
The number of all pairs of players is $\frac{25 \cdot 24}{2} = 300$ and after each game 10 of them become impossible. Therefore at most $300 \div 10 = 30$ games are possible.

We shall prove that 30 games are possible. We denote the pairs of players by $(m, n)$, where $1 \leq m, n \leq 5$ are integers (in other words, we put them in a table $5 \times 5$).

In the game $i$, $1 \leq i \leq 5$, we put the five pairs with $m = i$ (i.e. those from the $i$-th row of the table). In the game $6 + 5k + i$, $0 \leq i \leq 4$, $0 \leq k \leq 4$, we set the pair $(m, n)$ such that $mk + n$ is congruent to $i$ modulo 5. It is clear that for any fixed values of $k, i, m$ there exists a unique $n$ such that $mk + n \equiv i \pmod{5}$. Thus we have one pair in each row, i.e. the pairs are five and they have not played in the first 5 games.

For every two pairs $(m, n)$ and $(m', n')$, $m' \neq m$, the numbers $k(m - m')$, $k = 0, 1, 2, 3, 4$, give different remainders modulo 5. Hence there exists a unique $k$ such that $k(m - m') \equiv n - n' \pmod{5}$. Equivalently, $km - n$ and $km' - n'$ have the same remainder $i$ modulo 5 and the numbers $k$ and $i$ determine the unique game in which the pairs $(m, n)$ and $(m', n')$ participate.