Each match results in both players receiving $\frac{1}{2}$ point. After the competition, it is found that half of each player's points were obtained from matches against the ten lowest-scoring players. (The ten lowest-scoring players obtained half of their points from matches against each other), find the number of participants in the competition.
13. Suppose there are $n$ players, then it is known that the $n$ players scored a total of $\frac{n(n-1)}{2}-1$ points, and the 10 "losers" scored $5 \frac{10 \times 9}{2}=45$ points among themselves. Since this is half of their total score, they scored a total of 90 points. The remaining $n-10$ players scored $\frac{(n-10) \times(n-11)}{2}$ points among themselves, which is also half of their total score. Therefore, they scored $(n-10)(n-11)$ points in total. Thus, we have $\frac{n(n-1)}{2}=90-(n-10)(n-11)$, which simplifies to $(n-16)(n-25)=0$. Solving this, we get $n_{1}=25, n_{2}=16$ (the latter is discarded).

At this point, if there were only 16 players, then there would only be 6 "losers," and their total score would be 30, averaging 5 points per person, which is clearly less than the average score of the "losers," $\frac{90}{10}=9$. Therefore, $n=25$, meaning there are 25 players in the competition.