In a tournament, each participant plays a match against each other participant. The winner of a match earns 1 point, the loser 0, and if the match is a draw, both players earn half a point. At the end of the tournament, the participants are ranked according to their score (if multiple participants have the same score, their order is chosen randomly). Each participant won half of their points in their matches against the last ten participants (in the ranking). How many people participated in this tournament?
We will use a double counting method. By counting the number of points distributed in the tournament in two different ways, we will obtain an algebraic relationship involving the number $n$ of participants, which will allow us to conclude. A match corresponds to a distribution of 1 point, and thus the total score obtained by $k$ players in the matches they play against each other is $k(k-1) / 2$. Therefore, the total score of all participants is $n(n-1) / 2$.

The last 10 participants scored 45 points playing against each other, so their total score is 90 points. The other $n-10$ participants scored $(n-10)(n-11) / 2$ points playing against each other, and thus a total of $(n-10)(n-11)$ points. The total score of all participants is therefore $90+(n-10)(n-11)$ points. By comparing these two values, we obtain the equation $n^{2}-41 n+400=0$, which has two solutions: 16 and 25.

The tournament could not have contained 16 participants. Indeed, in this case, the total score of the last 10 participants is 90, and that of the first 6 is 30. The last 10 participants would therefore have scored on average more points than the first, which is absurd. The tournament could therefore have only 25 participants.