Let $P$ be one of the $20$ people. Let $F$ be the set of people among the other $19$ who are friends of $P$, and let $E$ be the set of those who are enemies of $P$; then $|F|+|E|=19$, since every one of the $19$ is either $P$’s friend or $P$’s enemy. If $|F|$ and $|E|$ were both less than $10$, then we’d have $$|F|+|E|\le 9+9=18\;,$$ which is impossible, so either $|F|\ge 10$ or $|E|\ge 10$. That is, either $P$ has at least $10$ friends in the group, or $P$ has at least $10$ enemies in the group.
Suppose that $P$ has at least $10$ friends in the group. Then $F$ contains either $3$ mutual friends or $4$ mutual enemies. If $F$ contains $4$ mutual enemies, obviously so does the whole group of $20$ people. If $F$ contains $3$ mutual friends, these $3$ people together with $P$ form a group of $4$ mutual friends.
The argument when $P$ has at least $10$ enemies is entirely similar.