In a club with 99 people, everyone knows at least 67 people. Prove there's a group of 4 people where everyone knows each other

You have shown that $A$ and $B$ have at least $35$ mutual acquaintances. Let $C$ be one of them. If $C$ knows any of the other $34$, then we have a group of four who all know each other. In order to avoid such a group, $C$ can know at most $98-34=64$ members (including $A$ and $B$), but it is given that $C$ knows at least $67$ members.

You can also use the Turan theorem. Suppose there is no such 4, then there is at most $${99^2\over 3}$$ acquaintances. But on the other hand we have by handshake lemma at least $${99\cdot 67\over 2}$$ acquaintances.

So we have $${99\cdot 67\over 2} \leq {99^2\over 3} \implies 67\cdot 3\leq 99\cdot 2$$

A contradiction!