Prove that any (2n+1)-element subset of{1,2, ...,3n}contains three consecutive integers.
My general thinking is that pigeonholes are sets of consecutive integers from {1,2,...,3n} which are {1,2,3}{4,5,6},...,{3n-2,3n-1,3n} and that each one is a pigeonhole filled 2/3 of the way, so the (2n+1)th element would have to fill a pigeonhole fully, but I don't know how to prove it.
You've got $n$ many holes, $\{1,2,3\},\{4,5,6\}, \ldots,\{3n-2,3n-1,3n\}$.
If you've got $2n+1$ balls to put in them, you cannot have at most $2$ per hole because that only leaves room for at most $2n$ balls, so one hole must contain $3$ and you're done.
So that partition indeed does work.