King Arthur's bodyguards held a knightly tournament. At the end of the tournament, it turned out that the king could find a third one for any two bodyguards who participated, who had defeated both of them. What is the minimum number of bodyguards who participated in the tournament?
Let's indulge in the idea that the king has at least two bodyguards, one of whom is $A$. This bodyguard $A$ was certainly replaced by someone, let's say $B$, and there was a bodyguard $C$ who defeated both $A$ and $B$. Now, $B$ could not have defeated $A$ and $C$, since $B$ was replaced by $C$, so this must have been a different bodyguard, say $D$. Bodyguard $A$ was defeated by $B$, $C$, and $D$, and since $A$ could be any bodyguard, every bodyguard must have lost at least three times.

![](https://cdn.mathpix.com/cropped/2024_05_02_e557b6445c78e790253bg-1.jpg?height=469&width=488&top_left_y=351&top_left_x=813)

Let's denote the number of knights participating in the tournament by $k$. In every match, one participant won and the other lost, and since every bodyguard lost at least three times, there were at least $3k$ matches. On the other hand, every knight fought with every other knight at most once, so

$$
3 k \leqq \frac{k(k-1)}{2}
$$

from which $k \leqq 7$, meaning at least 7 bodyguards participated in the tournament.

The table shows that the conditions described in the problem can be satisfied with 7 knights. No matter which two rows we choose, there is a column where both rows intersect with a star.