11. There are $n$ teams participating in a football league, each team plays a match with every other team only once. It is stipulated that a win earns 3 points, a draw earns 1 point, and a loss earns 0 points. After the league ends, some teams may be disqualified, and thus their match results will also be canceled. The remaining teams will have their scores recalculated, and the team with the most points will become the champion of this league (if only one team is not disqualified, then they are the champion).

Let $f_{i}(T)(i=1,2, \cdots, n)$ be the minimum number of teams that must be disqualified in this league $T$ to make the $i$-th team the champion. Let $F(T)=f_{1}(T)+f_{2}(T)+\cdots+f_{n}(T)$. For $n \geqslant 5$, find the maximum and minimum values of $F(T)$.
(53rd Belarusian Mathematical Olympiad (Class B))
Solution: Since $f_{1}(T) \leqslant n-1(i=1,2, \cdots, n)$, we have $F(T) \leqslant n(n-1)$. Additionally, in the league $T_{0}$, all matches end in a draw, so for the $i$-th team to become the champion, it must disqualify the other $n-1$ teams, hence $f_{1}\left(T_{0}\right)=n-1$. Therefore,
$$
F\left(T_{0}\right)=n(n-1).
$$

Thus, the maximum value of $F(T)$ is $n(n-1)$.
Next, we prove that the minimum value of $F(T)$ is $n$.
For the league $T_{1}$: the $i$-th team plays against the $(i+1)$-th team $(i=1,2, \cdots, n-1)$, and the $n$-th team plays against the 1st team, with all other matches ending in a draw. Clearly, disqualifying the $(i-1)$-th team allows the $i$-th team to win the championship $(i=2,3, \cdots, n)$, and disqualifying the $n$-th team allows the 1st team to win the championship, so $f_{1}\left(T_{1}\right)=1$, thus $F\left(T_{1}\right)=n$.

If in the league $T$ no team has more points than any other team, then $f_{i}(T) \geqslant 1(i=1,2, \cdots, n)$, hence $F(T) \geqslant n$;

If in the league $T$ one team has more points than any other team, let's assume the $n$-th team has more points than the others, then $f_{n}(T)=0, f_{i}(T) \geqslant 1(i=1,2, \cdots, n-1)$. Therefore, $F(T) \geqslant n-1$.

If $F(T)=n-1$, then $f_{i}(T)=1(i=1,2, \cdots, n-1)$, $f_{n}(T)=0$. Suppose the $i$-th team wins the championship, and the team disqualified is the $a_{i}$-th team, then when $i \neq j$, $a_{i} \neq a_{j}$. Thus, the $a_{i}$-th team must have lost to the $n$-th team (if $a_{i} \neq n$). Therefore, the $n$-th team must have won at least $n-2$ matches. Thus, the $n$-th team scores $k_{n} \geqslant 3(n-2)$ points, and the $i$-th team $(i=1,2, \cdots, n-1)$ scores $k_{i} \geqslant k_{n}-2$ points. Therefore, in the league $T$, the total points of all teams are
$$
\begin{array}{l}
\sum_{i=1}^{n} k_{i}=\sum_{i=1}^{n-1} k_{i}+k_{n} \\
\geqslant \sum_{i=1}^{n-1}\left(k_{n}-2\right)+k_{n}=n k_{n}-2(n-1) \\
\geqslant 3(n-2) n-2(n-1)=3 n^{2}-8 n+2 . \\
\text { On the other hand, } \sum_{i=1}^{n} k_{i} \leqslant 3 \cdot \frac{n(n-1)}{2}, \text { so, } \\
3 \cdot \frac{n(n-1)}{2} \geqslant 3 n^{2}-8 n+2 .
\end{array}
$$

This inequality does not hold when $n \geqslant 5$, leading to a contradiction.
Therefore, the minimum value of $F(T)$ is $n$.