Answer 5

The answer is $5$. To show that $n\le 4$ impossible, consider $\{0,3,4,7\}$. Now we show that $n=5$ works. We distinguish cases based on the distribution of the $5$ numbers among the residue classes modulo $3$. If three numbers are in the same residue class, then the sum of the three corresponding numbers is divisible by $3$. If none of the five numbers belong to a particular residue class, then by the Pigeonhole Principle, at least $\lceil\frac{5}{2}\rceil=3$ numbers must belong to one of the other two classes, and the sum of these is divisible by $3$. If there is one or two of the $5$ numbers in each residue class, we choose one number from each residue class, and their sum is divisible by $3$.