Interesting there are about $5$ in $10000$ chance that such a,b,c and d works

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Consider pairs $(a,b)$ and their sum $a+b$. There are $\binom{64}{2}=2016$ pairs and therefore sums. Let $a+b=c+d (mod\ | 2013)$ for some pairs $(a,b)$ , $(c,d)$ if they are distinct we are done , else $a=c$, then $b=d\ \ mod\ \ 2013$ and $b=d$ (all numbers are less than $2013$) so $(a,b)=(c,d)$ contradiction. We have $2016$ pairs and $2013$ possible residues so two pairs give same residue and we are done.