Oh man this problem is completely hilarious. There exists either $|A|=|B|=|C|=2$ or $|A|=|B|=|C|=3$ in fact. OK, so we look at $3$ element subsets, and note that $(68)(67)(11)>(8)((2015+2014+2013)-(1+2+3)+1)$, so it follows some $9$ three element subsets of $S={1,2,3,...2015}$ have equal sum by pigeonhole. Now, take these $9$ sets and make them vertices of a graph, connecting two vertices iff they do not share any elements. Note two vertices either share one or zero elements, because due to the same sum condition if they share two elements they also share a third, and are not different. If there is a triangle in this graph, we have found ${a,b,c}$, ${d,e,f}$, ${g,h,i}$ disjoint with $a+b+c=d+e+f=g+h+i$ and we are done. If not, by Turan's theorem it has at most $20$ edges, with the $K_{4,5}$ being an equality case. So therefore the average degree is at most $\frac {40} {9}$, and there is a vertex with degree at most $4$. Call this $V$, and say it is not adjacent to $V_1$, $V_2$, $V_3$, $V_4$. Say $V_0=V={a,b,c}$ and note that $V$ must share at least one of its elements, WLOG $a$, with at least two of the $V_i$ (WLOG $V_1$ and $V_2$). So then $V$, $V_1$, $V_2$ share an element $a$ and clearly they cannot share another element, because the same sum condition makes them equal. But then consider $W_i$ to be the two element set consisting of the elements of $V_i$ other than $a$ for $i=0$, $i=1$, $i=2$ and it's clear to see $W_0$, $W_1$, $W_2$ satisfy the given property so we are done in this case also.