China Girls' Mathematical Olympiad

Note that if there are $2m-4$ players, we can label them
$$a_1, a_2, \dots, a_{m-3}, A_{m-2}, B_{m-2}, a_{m-1}, \dots, a_{2m-5},$$
and assume that player $P_i$ beats player $P_j$ if and only if $i > j$, and $A_{m-2}$ and $B_{m-2}$ are in a tie. It is easy to see that in the group of $m$ players, there exists a unique player $P_i$ with the maximum index $i$ ($m-1 \le i \le 2m-5$, and this player won all games against other players in the group), and there exists a unique player $P_i$ with the minimum index $j$ ($1 \le i \le m-3$, and this player lost all games against other players in the group). Hence this tournament has property $P(m)$ and not all players have distinct total points. If $n < 2m-4$, we can then build a similar tournament by taking players away from both ends index-wise). Hence the answer is greater than $2m-3$. It suffices to show the following claim:
If there are $2m-3$ players in a tournament with property $P(m)$, then the players must have distinct total final score.

In a group, if a player won (or lose) all games against the rest of the players in the group, we call this player the winner (or loser) of the group. If a player won (or lose) all his games in the tournament, we call this player the complete winner (or complete loser). We establish the following lemmas.

Lemma 1 In an $n$-player ($n \ge m$) tournament with property $P(m)$, there is a complete winner.

Proof: We implement an induction on $n$. If $n=m$, the statement is trivial. Now assume that the statement is true for some $n=k$ ($k \ge m$), we consider a ($k+1$)-player tournament with property $P(m)$. Let $a_1, \cdots, a_{k+1}$ denote the players. By the induction hypothesis, we may assume that $a_{k+1}$ is the winner in the group $a_2, \cdots, a_{k+1}$. We consider three cases:
(a) If $a_{k+1}$ won the game against $a_1$, then $a_{k+1}$ is the complete winner;
(b) If $a_{k+1}$ tied the game against $a_1$, then the group $a_1, a_2, \cdots, a_{k-1}, a_{k+1}$ has no winner, violating the condition that the tournament has property $P(m)$;
(c) If $a_{k+1}$ lose the game against $a_1$, then the group $\{a_1, a_2, \cdots, a_{k-1}, a_k, a_{k+1}\} \setminus \{a_i\} (2 \le i \le k)$ has a winner, and this winner can only be $a_1$. Thus $a_1$ is the complete winner.
Combining the three cases, we find a complete winner in the tournament, hence our induction is complete.

In exactly the same way, we can prove that

Lemma 2 In an $n$-player ($n \ge m$) tournament with property $P(m)$, there is a complete loser.

Now we are ready to prove our claim in a similar manner.