8. $n$ chess players participate in a chess tournament, with each pair of players competing in one match. The rules are: the winner gets 1 point, the loser gets 0 points, and in the case of a draw, both players get 0.5 points. If it is found after the tournament that among any $m$ players, there is one player who has won against the other $m-1$ players, and there is also one player who has lost to the other $m-1$ players, this situation is said to have property $P(m)$.

For a given $m(m \geqslant 4)$, find the minimum value of $n$, denoted as $f(m)$, such that for any tournament situation with property $P(m)$, all $n$ players have distinct scores.
(Wang Jianwei provided)
8. First, prove two lemmas.

Lemma 1 When $n \geqslant m$, if the match results of $n$ players have property $P(m)$, then there must be a player who wins all others.
Proof of Lemma 1: When $n=m$, the proposition is obviously true.
Assume the proposition holds for $n$.
Then for $n+1$ players, choose any $n$ players. By the induction hypothesis, there must be a player who wins all others among these $n$ players. Without loss of generality, assume $A_{1}$, $A_{2}, \cdots, A_{n}$ where $A_{1}$ wins all others.
For the other player $A_{n+1}$:
If $A_{1}$ wins $A_{n+1}$, then among the $n+1$ players, $A_{1}$ wins all others;
If $A_{1}$ draws with $A_{n+1}$, consider the players $A_{1}, A_{2}, \cdots, A_{n-1}, A_{n+1}$, among these $n$ players, no one wins all others, which is impossible;

If $A_{n+1}$ wins $A_{1}$, consider the players $A_{1}, A_{3}, A_{4}, \cdots, A_{n}$, $A_{n+1}$, the one who wins all others can only be $A_{n+1}$, i.e., $A_{n+1}$ wins $A_{3}, A_{4}, \cdots, A_{n}$. Similarly, $A_{n+1}$ also wins $A_{2}$. Therefore, among these $n+1$ players, $A_{n+1}$ wins all others.
By the principle of induction, the proposition holds for any $n \geqslant m$.
Similarly, we can prove:
Lemma 2 When $n \geqslant m$, if the match results of $n$ players have property $P(m)$, then there must be a player who loses to all others.
Back to the original problem.
Next, prove: When $n \geqslant 2 m-3$, the scores of all players must be different.

By Lemma 1, there is a player $A_{1}$ who wins the other $n-1$ players, there is a player $A_{2}$ who wins the other $n-2$ players except $A_{1}$, $\qquad$ there is a player $A_{n-m+1}$ who wins $m-1$ players except $A_{1}, A_{2}, \cdots, A_{n-m}$.

By Lemma 2, there is a player $A_{n}$ who loses to the other $n-1$ players, there is a player $A_{n-1}$ who loses to the other $n-2$ players except $A_{n}$, $\cdots \cdots$ there is a player $A_{n-m+3}$ who loses to $n-m+2$ players except $A_{n}, A_{n-1}, \cdots, A_{n-m+4}$, and there is another player $A_{n-m+2}$.

Thus, among these $n$ players $A_{1}, A_{2}, \cdots, A_{n}$, the players with smaller numbers defeat those with larger numbers, so their scores are $n-1$, $n-2$, $\cdots$, $1$, $0$, all different.
For $n=2 m-4$, assume the skill levels of $n$ players are
$$
1,2, \cdots, m-3, m-2, m-2, m-1, \cdots, 2 m-5 \text {, }
$$
where players with smaller numbers defeat those with larger numbers, and players with the same number draw. Then, for any $m$ players chosen, there must be a minimum number $i(1 \leqslant i \leqslant m-3)$, and a maximum number $j(m-1 \leqslant j \leqslant$ $2 m-5)$, thus, among these $m$ players, the player with number $i$ wins all others, and the player with number $j$ loses to all others.

Therefore, these $n$ players have property $P(m)$, but two of them have the same score.
$$
\text { In summary, } f(m)=2 m-3 \text {. }
$$