6・39 In a chess tournament, each pair of players plays exactly one game, with the winner getting 1 point, the loser getting 0 points, and each player getting $\frac{1}{2}$ point in the case of a draw. After the tournament, it was found that each player's score was exactly half from games against the 10 lowest-scoring players (the 10 lowest-scoring players got half of their points from games against each other). Find the total number of players in the tournament.
[Solution] Let there be $n$ players participating in the competition. We refer to the 10 players with the lowest scores as "losers," and the remaining $n-10$ players as "winners."

According to the given information, the total number of matches played among the $n$ players is $C_{n}^{2}=\frac{1}{2} n(n-1)$, so the total score of the $n$ players is $\frac{1}{2} n(n-1)$.

Next, we calculate the total scores of the losers and the winners separately. The 10 losers play matches among themselves and score a total of $C_{10}^{2} = 45$ points. Since this is half of their total score, their total score is 90.
The $n-10$ winners play matches among themselves and score a total of
$$
C_{n-10}^{2}=\frac{1}{2}(n-10)(n-11),
$$

which is also half of their total score, so their total score is $(n-10)(n-11)$. Therefore, we have
$$
\begin{array}{c}
\frac{1}{2} n(n-1)=90+(n-10)(n-11), \\
n^{2}-41 n+400=0 .
\end{array}
$$

Solving this, we get $n_{1}=25, n_{2}=16$.
If there are 16 players in total, then there are 6 winners. Their total score is $2 C_{6}^{2} = 30$, averaging 5 points per person. However, the 10 losers average 9 points per person, which is impossible. Therefore, the total number of players participating in the competition is 25.