While reviewing some Olympiad training material I was stumped by this question.
Twenty people sit around a circular table for a discussion. After a break they again sit around the table, but in a different order. Prove that there are at least two people such that the number of participants sitting between them before and after the break is the same.
I cannot decide whether this is a combinatorics or pigeon hole question.
The phrase "number of participants sitting between them" is ambiguous to me, but I think it means that any permutation preserves the relative position of some elements.
I played around with this by hand and still cannot come up with a permutation in which this is not the case.
In other words, you want to prove that you can rotate the new seating position such that at least two people sit in their original position.
Label the chairs $1$ to $20$ clockwise. Say that person #$i$ sat in chair #$x_{i}$ before and in chair #$y_{i}$ after. This person will sit in his/her original position after $y_{i}-x_{i}$ rotation clockwise so all you need to prove is that not all $y_{i}-x_{i}$ are unique.
Hint: $\sum{x_{i}}$ equals $\sum{y_{i}}$ in modulo $20$