The question is as follows, it came in RMO in 1990. It most likely involves probability..as far I think.
Two boxes contain between them 65 balls of several different sizes. Each ball is white, black, red or yellow. If you take any five balls of the same colour, at least two of them will always be of the same size (radius). Prove that there are at least three balls which lie in the same box, have the same colour and the same size (radius).
$65$ balls and $4$ colours, so one colour identifies at least $17$ balls. Ignore all other balls apart from these with the $17$+ colour.
Regard these $17$+ as single element sets of balls. Take $5$ balls from the $17$+, and identify the pair of the same radius by merging their sets and reducing the number of sets to one less than before, so balls in the same set ar known to have the same colour and radius. Then repeat by again taking $5$ balls, each one from different sets, identifying the pair of the same radius, merging the sets the pair come from to reduce the number of sets by $1$, until there are only $4$ sets.
$17$+ balls of the same colour and $4$ equal-radius sets, so one set has at least $5$ balls of the same colour and radius.
$5$+ balls of the same colour and radius distributed between $2$ boxes requires at least one box to have at least $3$ balls of the same colour and radius.