I got this problem from Rustan Leino, who had heard several versions of it. He first heard it from Bertrand Meyer, who got the problem from Yuri Gurevich.
I solved it and wrote up my solution.
You're given a regular deck of 52 playing cards. In the pile you're given, 13 cards are face-up and the rest are face-down. You are to separate the given cards into two piles, such that the number of face-up cards in each pile is the same. In separating the cards, you're allowed to flip cards over. The catch: you have to do this in a dark room where you cannot determine whether a card is face-up or face-down.
All you have to do is separate the cards into a 13-card pile and a 39-card pile, then flip over all cards in the 13-card pile.
The reason this works is as follows. Let $n$ be the number of face-up cards in the 13-card pile when you first separate the cards into two piles. Because there are 13 face-up cards total between the two piles, the number of face-up cards in the other pile will be $13-n$. Now, when you flip over the 13-card pile, the $n$ face-up cards will become face-down and the remaining $13-n$ cards you flip will become face-up. Thus, both piles will wind up with the same number of cards face-up: $13-n$.