This is an old but good puzzle, which I personally enjoyed solving quite a while ago.

Seats of the round table are all labeled with the names of *N* managers. All managers have chosen the wrong seat. Prove that you can always rotate the table so that at least two managers sit in the positions with the correct names.

This problem is quite simple, so I don’t provide the solution, but feel free to share it in comments. As a hint, I can say that the whole solution shouldn’t be longer that one or two sentences.

This rotation is possible if at least two managers are in the same shifts from their correct seats. So it is impossible if all the managers are in different shifts. That means the 1st manager must be in the shift of 1, the 2nd – of 2 and so on to the last Nth manager who must be in the shift of N which is a total number of seats so he does the full circle and stays on his correct seat. By condition all the managers are on the wrong seats, so all the managers can’t be in different shifts from their correct seats. So you can always rotate the table in a given way

Dan, yes is more or less the solution which I came up with myself. My solution is phrased slightly different though. You can always put any single manager to a correct position by rotating a table. There are n managers in wrong positions, but there are only n-1 non-trivial rotations, so there must be a rotation, which puts two managers in the correct place.