There are $n$ football teams participating in a league, where each team plays a match against every other team exactly 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 that team is 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 ensure that the $i$-th team becomes the champion, and let $F(T)=f_{1}(T)+f_{2}(T)+\cdots+f_{n}(T)$. For $n \geqslant 5$, find the maximum and minimum values of $F(T)$.
Four Because $f_{i}(T) \leqslant n-1(i=1,2, \cdots, n)$, so $F(T) \leqslant n(n-1)$. In addition, in the league $T_{0}$, all matches are draws, then the $i$-th team must disqualify the other $n-1$ teams to win the championship, so $f_{i}\left(T_{0}\right)=n-1$. Thus, $F\left(T_{0}\right)=n(n-1)$.
Therefore, 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 wins the $i+1$-th team $(i=1,2, \cdots, n-1)$, the $n$-th team wins the 1st team, and all other matches are draws. Clearly, disqualifying the $i-1$-th team can make the $i$-th team the champion $(i=2,3, \cdots, n)$, and disqualifying the $n$-th team can make the 1st team the champion, so $f_{i}\left(T_{1}\right)=1$, thus $F\left(T_{1}\right)=n$.
If in the league $T$ no team has more points than the others, then $f_{i}(T) \geqslant 1(i=1,2, \cdots, n)$, therefore $F(T) \geqslant n$.
If in the league $T$ one team has more points than the others, without loss of generality, assume the $n$-th team has more points than the others, then $f_{n}(T)=0, f_{i}(T) \geqslant 1(i=1,2, \cdots, n-1)$. So $F(T) \geqslant n-1$.
If $F(T)=n-1$, then $f_{i}(T)=1(i=1,2, \cdots, n-1), f_{n}(T)=0$. Let the $i$-th team win 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 $a_{i} \neq n$). So the $n$-th team must have won at least $n-2$ matches. Thus, the team gets $k_{n} \geqslant 3(n-2)$ points, and the $i$-th team $(i=1,2, \cdots, n-1)$ gets $k_{i} \geqslant k_{n}-2$ points. Therefore, in the league $T$, the total points of all teams are
$$
\begin{aligned}
\sum_{i=1}^{n} k_{i} & =\sum_{i=1}^{n-1} k_{i}+k_{n} \\
& \geqslant \sum_{i=1}^{n-1}\left(k_{n}-2\right)+k_{n}=n k_{n}-2(n-1) \\
& \geqslant 3(n-2) n-2(n-1)=3 n^{2}-8 n+2 .
\end{aligned}
$$

On the other hand, $\sum_{i=1}^{n} k_{i} \leqslant 3 \cdot \frac{n(n-1)}{2}$, so
$$
3 \cdot \frac{n(n-1)}{2} \geqslant 3 n^{2}-8 n+2 .
$$

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