One, (20 points) In a certain competition, each player plays exactly one game against every other player. The winner of each game gets 1 point, the loser gets 0 points, and in the case of a draw, both get 0.5 points. After the competition, it is found that each player's score is exactly half from games played against the 10 lowest-scoring players (the 10 lowest-scoring players each have exactly half of their points from games played against each other). Find the number of participants in the competition.
Given $n$ players, the $n$ players scored a total of $\frac{n(n-1)}{2}$ points. The 10 "math players" scored a total of $\frac{10 \times 9}{2}=45$ points through their matches with each other, which is half of their total score, so they scored a total of 90 points. The remaining $n-10$ players scored a total of $\frac{(n-10)(n-11)}{2}$ points through their matches with each other, which is also half of their total score, so they scored a total of $(n-10)(n-11)$ points. Therefore,
$$
\frac{n(n-1)}{2}=90+(n-10)(n-11) \text {, }
$$

which simplifies to $(n-16)(n-25)=0$.
Solving this, we get $n_{1}=25, n_{2}=16$.
If there are 16 players, then there are only 6 "winners". Their total score is 30 points, averaging 5 points per person, which is clearly less than the average score of the losers, $\frac{90}{10}=9$ points. Therefore, $n=25$.
Thus, there are 25 players in the competition.