14. There are 10 table tennis players participating in a round-robin tournament. The match results show that there are no draws, and among any 5 players, there is 1 player who wins against the other 4, and 1 player who loses to the other 4. Then the number of players who won exactly two matches is $\qquad$.
14.1.

It can be proven that under the given conditions, no two players have the same number of wins. Therefore, the number of wins for 10 players are 10 different numbers: $0,1, \cdots, 9$. Hence, the number of players who win exactly two games is 1.

If not, suppose there exist $A$ and $B$ with the same number of wins, and without loss of generality, assume $A$ beats $B$. Then, among the players who lose to $B$, there must be a player $C$ who beats $A$, otherwise, any player who loses to $B$ would also lose to $A$, and $A$ would have at least one more win than $B$ (the game where $A$ beats $B$), which contradicts the assumption that $A$ and $B$ have the same number of wins.

Thus, we find three players $A$, $B$, and $C$ such that $A$ beats $B$, $B$ beats $C$, and $C$ beats $A$.

For $A$, $B$, and $C$, we can add 2 more players, and among these 5 players, there must be one player who loses to the other 4, and this player cannot be any of $A$, $B$, or $C$, denoted as $D$.

Similarly, for $A$, $B$, and $C$, we add 2 more players (excluding $D$), and we can find one player who loses to the other 4, and this player cannot be any of $A$, $B$, $C$, or $D$, denoted as $E$.

Thus, among the 5 different players $A$, $B$, $C$, $D$, and $E$, no one wins against the other 4, which contradicts the given condition.