For which $n\in\{3900, 3901,\cdots, 3909\}$ can the set $\{1, 2, . . . , n\}$ be partitioned into (disjoint) triples in such a way that in each triple one of the numbers equals the sum of the other two?
Source: Czech-Polish-Slovak Match 2007-P5
Tags: combinatorics proposed, combinatorics
For which $n\in\{3900, 3901,\cdots, 3909\}$ can the set $\{1, 2, . . . , n\}$ be partitioned into (disjoint) triples in such a way that in each triple one of the numbers equals the sum of the other two?