This article is from the Puzzles FAQ, by Chris Cole firstname.lastname@example.org and Matthew Daly email@example.com with numerous contributions by others.
Four glasses are placed upside down in the four corners of a square
rotating table. You wish to turn them all in the same direction,
either all up or all down. You may do so by grasping any two glasses
and, optionally, turning either over. There are two catches: you are
blindfolded and the table is spun after each time you touch the
glasses. Assuming that a bell rings when you have all the glasses up,
how do you do it?
1. Turn two adjacent glasses up.
2. Turn two diagonal glasses up.
3. Pull out two diagonal glasses. If one is down, turn it up and you're done.
If not, turn one down and replace.
4. Take two adjacent glasses. Invert them both.
5. Take two diagonal glasses. Invert them both.
"Probing the Rotating Table"
W. T. Laaser and L. Ramshaw
_The Mathematical Gardner_,
Wadsworth International, Belmont CA 1981.
... we will see that such a procedure exists if and
only if the parameters k and n satisfy the inequality
k >= (1-1/p)n, where p is the largest prime factor
The paper mentions (without discussing) two other generalizations:
more than two orientations of the glasses (Graham and Diaconis)
and more symmetries in the table, e.g. those of a cube (Kim).