Consider a tournament with $799$ contestants. Each contestant plays against all other contestants exactly one; there are no draws. Prove that there exist two disjoint groups $A,B$, of $7$ contestants each, such that everyone in $A$ beats everyone in $B$.
Suppose we choose $B$ by choosing any $7$ contestants randomly. If we can show that the expected number of contestants we can put into $A$ is at least $7$, we will be done. But is that true?
Edit: missed one division, so $S$ is $6$, which is not enough.
Consider the following algorithm. Starts with the set of contestant $S$ equal to everyone. By the Pigeonhole Principle, there must exists a contestant $l \in S$ that loses against at least half of the players in $S$. Put $l$ in the set $B$. Now, let $W_l \subset S$ be the set of contestants in $S$ that wins against $l$. Repeat this algorithm using $S = W_l$ 6 other times, and by the end of this, we would have put $7$ members in $B$. In addition, all the members we put in $B$ loses against the remaining members in $S$. What is the size of $S$?
$S \ge \lfloor 799/2^7 \rfloor \ge 6$.
Thus, we can pick any $7$ members of the remaining $S$ as $A$ and we have our set.