I think I've seen this problem before... If X, Y, Z are trays, then XY:Z denotes the operation done on X, Y to Z (take one candy from each of X and Y and put them to Z). Take a random tray A that has some candies and collect as many candies as possible here. (As long as there are two trays other than A that has candies, we can always do an operation.) If all candies come to A, we're done. Otherwise, we have another tray B with candies and trays C, D without candies. (Remember that there are at least four trays.) Either A and B both have two candies, in which AB:C, AB:C suffices, or otherwise some of them has three candies or more by Pigeonhole Principle; WLOG it is A (otherwise swap the labels A and B). Now we will move one candy from B to A. We do the following operations: AB:C, AC:B, AC:D, BD:A, BD:A. It can be easily verified that B loses exactly one candy and A gains exactly one candy. The only conditions required are that A has at least three candies, which we have assured by the previous paragraph, and B has at least one candy. When B no longer has any candy, then all candies are on A and we're done.