Romanian Mathematical Olympiad

Consider the following sets, each containing 335 elements $$\begin{array}{rlrl}A& =\{6,12,\dots ,2010\},& B& =\{3,9,15,\dots ,2007\},\\ C& =\{1,7,13,\dots ,2005\},& D& =\{2,8,14,\dots ,2006\},\\ E& =\{4,10,16,\dots ,2008\},& F& =\{5,11,17,\dots ,2009\}.\end{array}$$ If $S$ contains two elements from $A$ or two elements from $B$, their sum is divisible by $6$. If not, the remaining $4$ sets contain at least $673-2=671$ elements from $S$.

Consider the sets $C\cup F$ and $D\cup E$, each containing $670$ elements. One of the intersections of $S$ with these, say $C\cup F$, contains at least $336$ elements. Thus $S\cap C$ and $S\cap F$ contain each at least one element. Their sum is a multiple of $6$, which is what was to be proven.