In a certain competition, each player plays exactly one game with every other player. The winner of each game gets 1 point, the loser gets 0 points. If it's a tie, each gets $\frac{1}{2}$ point. After the competition, it is found that each player's score is exactly half from the games played against the ten lowest-scoring players. (The ten lowest-scoring players got half of their points from games played against each other) Find the total number of players in the competition.
Let's assume there are $n$ participants in the competition, and we will find two quantities related to $n$: one is the total points of the scores, and the other is the total points of the "losers" (let's call this sum). We need to note that if $k$ participants compete, the total number of matches should be $\frac{k(k-1)}{2}$, and according to the scoring rules, their total score is $\frac{k(k-1)}{2}$ points.

From this, we know that $n$ participants have a total score of $\frac{n(n-1)}{2}$ points. The 10 "losers" have scored a total of $10 \cdot 9 / 2 = 45$ points through their matches with each other. Since this is half of their total score, they have a total of 90 points. The remaining $n-10$ participants have scored a total of $(n-10)(n-11) / 2$ points through their matches with each other. This is also half of their total score, so they have a total of $(n-10)(n-11)$ points. Therefore, we get the equation
$$
\mathbf{n}(\mathbf{n}-1) / 2 = 90 + (\mathbf{n}-10)(\mathbf{n}-11),
$$

which is equivalent to $n^2 - 41n + 400 = 0$. That is,
$$
(n-16) \quad (n-25) = 0.
$$
$n_1 = 25, n_2 = 16$. Upon verification, $n_2 = 16$ should be discarded. Because if there are only 16 participants, there would only be 6 "winners," and their total score would be 30, averaging 5 points per person, which is clearly less than the average score of the "losers," $90 / 10 = 9$. Therefore, $n = 25$, and there were 25 participants in the competition.