A short while ago I came across this quite famous Russian problem which has a very intriguing result, but I myself have made little headway on it thus far:
In a chess tournament there are \( 2n+3 \) competitors. Every competitor plays every other competitor precisely once. No 2 matches can be played simultaneously, and after a competitor plays a match he can not play in any of the next n matches. Show that one of the competitors who plays in the opening match will also play in the closing match.
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Here's an initial thought. The first 2 players cannot play in the next n games, which involve 2n more people. Hence, the next game after that must involve the last person, and one of the first 2 players.
Can we continue from here?