Four (This sub-question is worth 50 points)
$n$ chess players participate in a chess tournament, with each pair of players competing in one match. The winner gets 1 point, the loser gets 0 points, and in the case of a draw, both get 0.5 points. If it is found after the tournament that among any $m$ players, there is one player who has defeated 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 $m (m \geq 4)$, find the minimum value of $n$, denoted as $f(m)$, such that for any situation with property $P(m)$, all $n$ players have different scores.

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Proof: We first prove two lemmas.

Lemma 1 When $n \geq m$, if the match results of $n$ players have property $P(m)$, then there must be a player who wins all their matches.
When $n=m$, the proposition is obviously true.
Assume the proposition holds for $n$, then for $n+1$ players, take any $n$ players. By the induction hypothesis, there must be a player who wins all their matches among these $n$ players, let's say $A_{1}, A_{2}, \ldots, A_{n}$, where $A_{1}$ wins all.
For the other player $A_{n+1}$:
If $A_{1}$ wins $A_{n+1}$, then among the $n+1$ players, $A_{1}$ wins all;
If $A_{1}$ draws with $A_{n+1}$, consider the players $A_{1}, A_{2}, \ldots, A_{n-1}, A_{n+1}$, there is no one who wins all, which is impossible; if $A_{n+1}$ wins $A_{1}$, consider the players $A_{1}, A_{3}, A_{4}, \ldots, A_{n}, A_{n+1}$, the one who wins all can only be $A_{n+1}$, especially, $A_{n+1}$ wins $A_{3}, A_{4}, \ldots, A_{n}$. Similarly, $A_{n+1}$ also wins $A_{2}$, so among these $n+1$ players, $A_{n+1}$ wins all.
By the principle of induction, the proposition holds for any $n \geq m$.

Similarly, we can prove:
Lemma 2 When $n \geq m$, if the match results of $n$ players have property $P(m)$, then there must be a player who loses all their matches.

Back to the original problem. We will prove:
When $n \geq 2 m-3$, 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 $A_{1}$, ..., there is a player $A_{n-m+1}$ who wins the other $m-1$ players except $A_{1}, A_{2}, \ldots, A_{n-m}$.

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

Thus, these $n$ players $A_{1}, A_{2}, \ldots, A_{n}$, the players with smaller numbers defeat the players with larger numbers, so their scores are $n-1, n-2, \ldots, 1, 0$, all different.
For $n=2 m-4$, assume the levels of the $n$ players are:
$$
1,2, \ldots, m-3, m-2, m-2, m-1, \ldots, 2 m-5,
$$

where players with smaller level numbers win players with larger level numbers, and players with the same level number draw. Then for any $m$ players, there must be a minimum level number $i(1 \leq i \leq m-3)$, and another maximum level number $j(m-1 \leq j \leq 2 m-5)$, thus among these $m$ players, the player with level $i$ wins all, and the player with level $j$ loses all, so these $n$ players have property $P(m)$, but there are two players with the same score.
In summary, $f(m)=2 m-3$.