Danube Mathematical Competition

Using the 64 given numbers, there can be formed 2016 pairs $(a, b)$, with $a < b$; these pairs generate 2016 sums $a+b$ which give 2016 remainders when divided by 2013. Consequently, there are two different pairs $(a, b)$ and $(c, d)$ having the same remainder when divided by 2013, hence $2013 \mid (a+b) - (c+d)$.

Assuming that $(a, b)$ and $(c, d)$ have a common component, e.g. $a = c$, then $2013 \mid b - d$, and, since $|b-d| \le 2012$, we get $b = d$, contradiction. The conclusion now follows.