Play a match. The winning team gets 3 points, the losing team gets 0 points, and in case of a draw, each team gets 1 point. How many points must a team score at least to ensure that no more than $k-1$ teams score no less than this team, where $2 \leqslant$ $k \leqslant n-1$?
Solution: Clearly, the worst-case scenario is that $k+1$ teams have the same score and all have the highest score.
(1) When $k$ is even, let $k=2 m$. Represent the $2 m+1$ teams with $2 m+1$ equally spaced points on a circle. Each team defeats the $m$ teams that follow it in a clockwise direction and loses to the other $m$ teams. At the same time, these $k+1$ teams each defeat all the other $n-k-1$ teams. Thus, the score of each of these $k+1$ teams is
$$
3(n-k-1)+3 m=3 n-\frac{3}{2} k-3 \text{. }
$$
(2) When $k$ is odd, let $k=2 m+1$. Still represent the $2 m+2$ teams with $2 m+2$ equally spaced points on a circle. As in (1), each team defeats the $m$ teams that follow it in a clockwise direction, draws with the $m+1$-th team, and loses to the other $m$ teams. In addition, these $k+1$ teams each defeat all the other $n-k-1$ teams. Therefore, the score of each of these $k+1$ teams is
$$
3(n-k-1)+3 m+1=3 n-\frac{1}{2}(3 k+1)-3 \text{. }
$$

Combining (1) and (2), whether $k$ is odd or even, the score of the $k+1$ teams is the same, which is $3 n-\left[\frac{3 k+1}{2}\right]-3$. This indicates that when a team's score is $3 n-\left[\frac{3 k+1}{2}\right]-3$, it is not enough to ensure that there are at most $k-1$ teams with a score no less than this team.

Next, we prove that when a team's score is no less than $3 n-\left[\frac{3 k+1}{2}\right]-2$, the number of teams with a score no less than this team is at most $k-1$.

If not, suppose there are $k$ teams with a score no less than $3 n-\left[\frac{3 k+1}{2}\right]-2$. Then, the total score of these $k+1$ teams is no less than
$$
(k+1)\left(3 n-\left[\frac{3 k+1}{2}\right]-2\right) \text{. }
$$

Let these $k+1$ teams be Group A, and the other $n-k-1$ teams be Group B. The total score of the matches between Group A and Group B is at most
$$
3(k+1)(n-k-1) \text{. }
$$
The total score of the matches within Group A is at most
$$
3 \mathrm{C}_{k+1}^{2}=\frac{3}{2} k(k+1) \text{. }
$$

Therefore, the total score of Group A is no more than
$$
\begin{aligned}
& 3(k+1)(n-k-1)+\frac{3}{2} k(k+1) \\
= & (k+1)\left(3 n-3 k-3+\frac{3}{2} k\right) \\
= & (k+1)\left(3 n-\frac{3}{2} k-3\right) \\
< & (k+1)\left(3 n-\left[\frac{3 k+1}{2}\right]-2\right) \text{. }
\end{aligned}
$$

Contradiction.
In summary, the minimum score required is
$$
3 n-\left[\frac{3 k+1}{2}\right]-2 \text{. }
$$