There are $281 n(n>3)$ football teams that have played $k$ matches, with each pair of teams playing at most one match, and there are no ties. It turns out that exactly one team remains undefeated. Additionally, it is found that for any two teams $A$ and $B$, there exists another team $C$ such that either $A$ beats $C$ and $B$ beats $C$; or $C$ beats $A$ and $C$ beats $B$. Find the minimum value of $k$, denoted as $f(n)$.
Solve $f(n)=2 n-2$.
Represent teams with points. If team $A$ beats team $B$, draw a line segment between $A$ and $B$ with an arrow pointing to $A$, forming a graph $G$.

Next, we prove: the minimum number of edges $k$ in a graph $G$ that meets the problem's requirements is $2 n-2$.

First, the graph in Figure 3 has exactly $2 n-2$ edges and meets the problem's requirements.

Second, for any graph $G$ that meets the requirements, we prove that the number of edges is no less than $2 n-2$.

We use mathematical induction on $n$.

When $n=4$, if the number of edges is less than 6, assume there is no edge between $A$ and $B$. Without loss of generality, assume the arrows from $A$ and $B$ all point to $C$ (as shown in Figure 4).

For $A$ and $C$, by the problem's conditions, there are only two possibilities. Considering $B$ and $D$, and $B$ and $C$, none of these cases meet the requirements. Therefore, the number of edges is no less than 6.
Assume the conclusion holds for $n-1$. Now consider the case for $n$.
If there is a point in the graph with a degree of 2, then the arrows pointing to and away from this point each have one (otherwise, the two endpoints of the edge connected to this point do not meet the requirements). Removing this point and the two edges connected to it, by the induction hypothesis, the remaining $n-1$ points have at least $2 n-4$ edges, totaling no less than $2 n-2$ edges. The conclusion holds.

If every point in the graph has a degree of at least 3, assume the number of edges $k=2 n-3$, and let $A_{0}$ be undefeated.

By the problem's conditions, there is no team that loses every game (otherwise, let it be $B$, then for $B$ and $A_{0}$, there is no team $C$ that meets the problem's requirements). Let the other $n-1$ teams be $A_{1}, A_{2}, \cdots, A_{n-1}$, and the degrees of the $n$ points be $a_{0}, a_{1}, \cdots, a_{n-1}$. Define an angle where the arrows from a point point in the same direction as a "same-direction angle," and an angle where the arrows point in different directions as a "different-direction angle."

By the problem's conditions, for any two teams, there is at least one same-direction angle, and the same-direction angles for different pairs of teams are different. Therefore, the graph has at least $\frac{n(n-1)}{2}$ same-direction angles.
By the adjustment method, it is easy to prove the following lemma.
Lemma If $x_{i} \in \mathbf{N}, x_{i} \geqslant m(i=1,2, \cdots, t)$, then
$x_{1} x_{2} \geqslant m\left(x_{1}+x_{2}-m\right)$,
$$
\sum_{i=1}^{t} x_{i}^{2} \geqslant(t-1) m^{2}+\left(\sum_{i=1}^{t} x_{i}-m t+m\right)^{2} \text {. }
$$

Returning to the original problem.
For any point $A_{i}(1 \leqslant i \leqslant n-1)$, let the number of arrows pointing away from and towards it be $x_{i}\left(x_{i} \geqslant 1\right)$ and $y_{i}\left(y_{i} \geqslant 1\right)$, respectively. Then $x_{i}+y_{i}=a_{i} \geqslant 3$.
The number of different-direction angles with it as the vertex is
$x_{i} y_{i}=x_{i}\left(a_{i}-x_{i}\right) \leqslant a_{i}-1$.
By the lemma, the total number of same-direction angles is
$$
\sum_{i=0}^{n-1} \frac{a_{i}\left(a_{i}-1\right)}{2}-\sum_{i=1}^{n-1} x_{i} y_{i}
$$
$\leqslant \sum_{i=0}^{n-1} \frac{a_{i}\left(a_{i}-1\right)}{2}-\sum_{i=1}^{n-1}\left(a_{i}-1\right)$
$=\frac{1}{2} \sum_{i=0}^{n-1} a_{i}^{2}-\frac{3}{2} \sum_{i=0}^{n-1} a_{i}+a_{0}+n-1$
$=\frac{1}{2} \sum_{i=0}^{n-1} a_{i}^{2}-3 k+a_{0}+n-1$
$\leqslant \frac{1}{2} \sum_{i=0}^{n-1} a_{i}^{2}-4 n+7$
$\leqslant \frac{1}{2}\left[(n-1) \times 3^{2}+\left(\sum_{i=0}^{n-1} a_{i}-3 n+3\right)^{2}\right]-4 n+7$
$=\frac{1}{2}\left[9(n-1)+(n-3)^{2}\right]-4 n+7$
$$
=\frac{n^{2}-n}{2}-(2 n-7)<\frac{n^{2}-n}{2} \text {, }
$$

Contradiction.
When $k<2 n-3$, adding several edges can transform it into the case where $k=2 n-3$.
Thus, the conclusion holds for $n$.
In summary, the minimum value is $f(n)=2 n-2$.