Must be $3|\frac{n(n+1)}{2}$or $n=0,-1\mod 3$ and $n\le \frac{n(n+1)}{6}$ or $n\ge 5$. If we find subsets for $n$ $A_n,B_n,C_n$, then we can find for $n+6$ as $A_{n+6}=\{A_n,n+1,n+6\},$ $B_{n+6}=\{B_n,n+2,n+5\}$, $C_{n+6}=\{C_n,n+3,n+4\}.$ If $n=5$, then $A_5=\{1,4\}, \ B_5=\{2,3\}, \ C_5=\{5\}$. It give solution for $n=5\mod 6$ If $n=6$, then $A_6=\{1,6\}, \ B_6=\{2,5\}, \ C_6=\{3,4\}.$ If $n=8$, then $A_8=\{1,2,3,6\}, \ B_8=\{4,8\}, \ C_8=\{5,7\}.$ If $n=9$, then $A_9=\{1,2,3,4,5\} \ B_9=\{6,9\}, C_9=\{7,8\}.$

What is special here of n+6 .??i am asking the motivation for considering n+6 .

Do we need to check like this ..if there is any period ??anyone