3 There are $n(n \geqslant 5)$ football teams participating in a league, where each pair of teams plays exactly one match. 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. Among the remaining teams, if a team's points exceed those of any other team, then that team becomes the champion of this league (if only one team is not disqualified, then it is the champion). Let the minimum number of teams that need to be disqualified to make the $i$-th team the champion be $f_{i}$. Find the maximum and minimum values of $F=f_{1}+\cdots+f_{n}$. (53rd Belarusian Mathematical Olympiad problem)
3. If all teams except team $i$ are disqualified, then team $i$ naturally becomes the champion, so $f_{i} \leqslant n-1$. Therefore, $F \leqslant n(n-1)$. Furthermore, when all teams draw, for all $i=1,2, \cdots, n$, we have $f_{i}=n-1$. At this point, $F=n(n-1)$. Hence, $F_{\text {max }}=n(n-1)$. When $n \geqslant 5$, we first prove that $F \geqslant n$. A sufficient condition for $F \geqslant n$ is $f_{i} \geqslant 1$, meaning no team has more points than all other teams. Additionally, assume that for $i=1,2, \cdots, n-1$, team $n$'s points $S_{n}$ are more than team $i$'s points $S_{i}$, then team $i$ must eliminate at least one team to become the champion, i.e., $f_{i} \geqslant 1$. (1) When $a_{i} = n$, team $i$ must win against team $n$ by at least 2 points. If team $i$ eliminates team $a_{i}$, team $n$'s points are at least $S_{n}-3$, while team $i$'s points do not increase. Since $S_{n}-3 \geqslant S_{i}$, this contradicts the requirement for team $i$ to be the champion. (2) When $a_{i} \neq n$, team $a_{i}$ loses to team $n$. Otherwise, after eliminating $a_{i}$, team $n$'s points are at least $S_{n}-1$, while team $i$'s points do not increase. Since $S_{n} > S_{i}$, it follows that $S_{n}-1 \geqslant S_{i}$, which contradicts the requirement for team $i$ to be the champion. (3) When $i \neq j$, $a_{i} \neq a_{j}$. Otherwise, after eliminating $a_{i}$, there would be two champion teams, team $i$ and team $j$, which is a contradiction. From (2) and (3), we know that all teams $a_{1}, a_{2}, \cdots, a_{n-1}$, except team $n$, lose to team $n$. Thus, at least $n-2$ teams lose to team $n$, so $S_{n} \geqslant 3(n-2)$. Therefore, $S=\sum_{i=1}^{n} S_{i}=\sum_{i=1}^{n-1} S_{i}+S_{n} \geqslant \sum_{i=1}^{n-1}\left(S_{n}-2\right)+S_{n}=(n-1)\left(S_{n}-2\right)+S_{n}=n S_{n}-2(n-1) \geqslant n \cdot 3(n-2)-2(n-1)=3 n^{2}-8 n+2$. On the other hand, each match contributes at most 3 points to $S$, so $S \leqslant 3 \mathrm{C}_{n}^{2}=3 \cdot \frac{n(n-1)}{2}$. Therefore, $3 \cdot \frac{n(n-1)}{2} \geqslant 3 n^{2}-8 n+2$, solving this gives $n<5$, which is a contradiction. Finally, when $A_{1}$ beats $A_{2}$, $A_{2}$ beats $A_{3}$, ..., $A_{n}$ beats $A_{1}$, and all other matches end in a draw, it can be verified that all $f_{i}=1$, at this point $F=n$. Thus, when $n \geqslant 5$, $F_{\min }=n$.