6.41 In a certain chess tournament, each player obtained half of their points from games played against the last 3 players. How many people participated in this tournament?

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[Solution] The last 3 players play 3 matches among themselves, scoring a total of 3 points. Since this is half of their total points, they must have scored another 3 points in matches against the top players.

Let there be $n$ players in total. Thus, the top $n-3$ players play $C_{n-3}^{2}$ matches among themselves, scoring a total of $C_{n-3}^{2}$ points. According to the given information, this is half of the points scored by the top $n-3$ players. The other half is from matches against the last 3 players.
The top $n-3$ players play a total of $3(n-3)$ matches against the last 3 players, scoring a total of $3(n-3)$ points, hence we have
$$
3(n-3)-3=C_{n-3}^{2}=\frac{1}{2}(n-3)(n-4) \text {. }
$$

Solving this, we get $n=9$.
When there are 9 players in the competition, the top 6 players can draw with each other, and the last 3 players can also draw in all 3 of their matches. In the matches between the top 6 and the last 3 players, players 1 and 2 win against 7 and 8, and draw with 9; players 3 and 4 win against 8 and 9, and draw with 7; players 5 and 6 win against 9 and 7, and draw with 8. Thus, each of the top 6 players scores 5 points, and each of the last 3 players scores 2 points, with half of their points coming from matches against the last 3 players.
In summary, there are 9 players in the competition.