National Olympiad of Argentina

Among 5 arbitrary integers either there are three with the same remainder modulo 3 or three with different remainders modulo 3. In both cases the sum of these three is divisible by 3.

If in addition the integers are positive and distinct like in a) then the sum in question is at least $1+2+3>3$, hence it is not a prime. So there can be at most 4 numbers that satisfy condition a), and 4 such numbers exist. There are many examples: $\{1,3,7,9\}$; $\{3,5,11,15\}$; $\{7,13,17,23\}$; $\{7,13,23,53\}$ etc.

Now let $x_1<x_2<\cdots <x_6$ be 6 integers satisfying b). Among their sums by triples consider $S_1=x_1+x_2+x_3$, $S_2=x_1+x_2+x_4$ and the sum $S$ of an arbitrary triple that does not contain $x_1$. Note that $S>S_2>S_1$. Since $S_1$ and $S_2$ are (positive) prime numbers, they are at least 2 and 3 respectively; hence $S>3$. However by the introductory remark there are three among the 5 numbers $x_2,x_3,\dots ,x_6$ with sum $S$ divisible by 3. Combined with $S>3$ this yields a contradiction with $S$ being a prime. It remains to show that there exist 5 numbers satisfying b). There is a variety of examples here too, with one or two negative numbers: $\{-13,-1,17,25,55\}$; $\{-11,-5,19,23,29\}$; $\{-9,-3,15,25,31\}$; $\{-9,3,9,19,229\}$ etc.