6.4 20 football teams participate in the national championship. How many matches must be played at least, so that in any 3 teams, there are always two teams that have played against each other?
[Solution] Suppose that after $m$ matches, any 3 teams have already played against each other. Let team $A$ be the team with the fewest matches played, having played $k$ matches. Thus, the teams that have played against $A$ have each played at least $k$ matches. The $19-k$ teams that have not played against $A$ must each play against each other, otherwise there would exist 3 teams where no two have played against each other. Therefore, we have
$$
2 m \geqslant(k+1) k+2 C_{19-k}^{2}=2(k-9)^{2}+180 \geqslant 180 .
$$

This implies that at least 90 matches must be played.
On the other hand, if the 20 teams are evenly divided into two groups, with every two teams within the same group playing against each other and no matches between teams from different groups, then a total of 90 matches are played. Since any 3 teams will always have two in the same group, and they have already played against each other, this arrangement meets the requirements of the problem.
In summary, the minimum number of matches to be played is 90.