1. In an international chess tournament with 10 participants, each player must play exactly one game against every other player. After several games, it is found that among any three players, at least two have not yet played against each other. How many games have been played at most by this point?
2. Suppose there are 5 men and 5 women among the 10 players, and all matches so far have been between men and women, thus satisfying the condition, i.e., 25 matches have been played.
Next, we prove: 25 is indeed the maximum value.
Let $k$ be the number of matches played by the player who has played the most matches, denoted as $A$. The $k$ players who have played against $A$ have not played against each other, meaning each of them has played at most 10 $k$ matches. The remaining $10-k$ players have each played at most $k$ matches. Therefore, the total number of matches is at most
$$
\begin{array}{c}
\frac{k(10-k)+(10-k) k}{2}=-k^{2}+10 k \\
=25-(k-5)^{2} \leqslant 25 .
\end{array}
$$