6.57 $n$ football teams are participating in a tournament, where every two teams play each other once. The winning team gets 3 points, the losing team gets 0 points, and in case of a draw, both teams get 1 point. How many points does a team need to score to ensure that at most $k-1$ teams score no less than this team, where $2 \leqslant k \leqslant n-1$?
[Solution] Consider the case where $k+1$ football teams have the same and highest score.
(1) Let $k$ be even, denoted as $k=2 m, m \in \mathbb{N}$. Represent the $2 m+1$ teams with $2 m+1$ equally spaced points on a circle. Each team wins against the next $m$ teams in the clockwise direction from its corresponding point and loses to the other $m$ teams. Additionally, each of these $k+1$ teams wins against the other $n-k-1$ teams. Thus, the score of each of these $k+1$ teams is $3(n-k-1)+3 m=3 n-3 k-3+\frac{3}{2} k=3 n-\frac{3}{2} k-3$.
(2) Let $k$ be odd, denoted as $k=2 m+1, m \in \mathbb{N}$. Similarly, represent the $2 m+2$ teams with $2 m+2$ equally spaced points on a circle. Each team wins against the next $m$ teams in the clockwise direction from its corresponding point, draws with the $(m+1)$-th team, and loses to the other $m$ teams. Additionally, each of these $k+1$ teams wins against the other $n-k-1$ teams. Thus, the score of each of these $k+1$ teams is
\[
\begin{aligned}
3(n-k-1)+3 m+1 & =3 n-3 k-3+\frac{3}{2}(k-1)+1 \\
& =3 n-\frac{3 k+1}{2}-3 .
\end{aligned}
\]

Combining the results from (1) and (2), the score of each of these $k+1$ teams is $3 n-\left[\frac{3 k+1}{2}\right]-3$. This implies that to ensure that at most $k-1$ teams score no less than a certain team, that team must score at least $3 n-\left[\frac{3 k+1}{2}\right]-2$ points.
(3) Now, prove that if a team scores no less than $3 n-\left[\frac{3 k+1}{2}\right]-2$ points, then at most $k-1$ teams score no less than this team.

Suppose otherwise, that there are $k$ other teams scoring no less than this team. Thus, the total score of these $k+1$ teams is at least $(k+1)\left(3 n-\left[\frac{3 k+1}{2}\right]-2\right)$.

Let these $k+1$ teams be Group A, and the other $n-k-1$ teams be Group B. It is clear that the total score from matches between Group A and Group B is at most $3(k+1)(n-k-1)$, and the total score from matches within Group A is at most $3 C_{k+1}^{2}=\frac{3}{2} k(k+1)$. Therefore, the total score of Group A is at most
\[
\begin{aligned}
& 3(k+1)(n-k-1)+\frac{3}{2} k(k+1) \\
= & (k+1)\left(3 n-3 k-3+\frac{3}{2} k\right) \\
= & (k+1)\left(3 n-\frac{3}{2} k-3\right)<(k+1)\left(3 n-\left[\frac{3 k+1}{2}\right]-2\right),
\end{aligned}
\]
which is a contradiction.
In conclusion, the minimum score required is $3 n-\left[\frac{3 k+1}{2}\right]-2$.