Example 1: 20 teams participate in the national football championship. To ensure that in any three teams that have already played, there are two teams that have already played against each other, what is the minimum number of matches that need to be played?
(3rd All-Union Mathematical Olympiad)
Solution: Suppose that in any three teams that have already played, there are two teams that have played against each other. Let team $A$ be the team with the fewest matches played, having played $k$ matches. Thus, the $k$ teams that have played against team $A$, as well as team $A$ itself, have all played no fewer than $k$ matches.

Among the $19-k$ teams that have not played against team $A$, each team has played against the other $18-k$ teams. Thus, in any three of these teams, they have all played against each other.
It is evident that twice the total number of matches is no less than
$$
\begin{array}{l}
k(k+1)+(19-k)(18-k) \\
=2 k^{2}-36 k+342 \\
=2(k-9)^{2}+180 \geqslant 180,
\end{array}
$$

which means the number of matches is no less than 90.
Now, let's construct an example: divide the 20 teams into two groups, with each group of 10 teams playing against each other, but teams from different groups do not play against each other. This schedule meets the requirements of the problem, meaning at least $2 \mathrm{C}_{0}^{2}=90$ matches are needed.