Basic Pigeonhole. Note that $\binom{65}{2}=2080>2016$. In particular, there exists two distinct $2-$element subsets of $A$ whose sums are congruent to each other in modulo $2016$ (by Pigeonhole). Let these pairs be $(a',b')$ and $(c',d')$. If they have an element in common, say, $a'=c'$, then $b'\equiv d'\pmod{2016}$, a contradiction. Thus, pairs have no element in common, thus we can use $(a,b,c,d)=(a',b',c',d')$ above.