$\bullet$ If $ n\equiv1 \pmod{3}$ then there are no such triple sets. The proof is easy by contradiction: If there are such sets as $(A,B,C)$ then the sum of all the numbers in these three sets is divisible by $3$ but we also know that these numbers are $\{ 1 , 2 , ... , n \}$ so their sum is $\frac{n(n+1)}{2}$ which is not divisible by $3$ so we have a contradiction. $\bullet$ If $ n\equiv0 \pmod{3}$ or $ n\equiv2 \pmod{3}$ there are $3^{n-1}$ such triples of sets because for any single number from the $\{ 2 , ... , n \}$ there is $3$ of which set to put it in ($A$ or $B$ or $C$) and there are $n-1$ numbers and because $2+..+n\equiv2 \pmod{3}$ then the sum of the sets $(A,B,C) \pmod{3}$ is either: Two $0$s and one $2$ so we have to add the number $1$ that is left to the set that is $2\pmod{3}$ Or two $1$s and one $0$ so we have to add the number $1$ that is left to the set that is $0\pmod{3}$ Or two $2$s and one $1$ so we have to add the number $1$ that is left to the set that is $1\pmod{3}$ So in all the options we can only put $1$ in a single set so that the sets are all equal mod $3$ and therefore there is only one choice in where to put $1$ and in all the options it is possible to put $1$ in a set so that they are all equal mod $3$