We need ${{n(n+1)}\over 2}\equiv 0(3)\iff n\equiv 0,2(3)$ which shows that it is impossible for $n=2014$. If it's possible for $n$ then also for $n+9$ since we can do $$A\rightarrow A\cup \{n+9,n+4,n+2\},B\rightarrow B\cup \{n+8,n+6,n+1\}, C\rightarrow C\cup \{n+7,n+5,n+3\}$$For $n=8$ we have $A=\{8,4\},B=\{7,5\},C=\{6,1,3,2\}$ hence we have a solution for $n\equiv 8(9)$ which includes 2015. For $n=11$ we have $A=\{11,10,1\},B=\{9,8,5\},C=\{7,6,4,3,2\}$ hence soluble for $n\equiv 11(9)$ which includes 2018.