There exists an integer with alternating digits $1$ and $2$ which is divisible by $2013$ Could someone give me hints in how to solve the following (rather interesting) problem? 

Prove that there exists an integer consisting of an alternance of $1$s
  and $2$s with as many $1$s as $2$s (as in $12$, $1212$, $1212121212$, etc.), and
  which is divisible by $2013$.

Source: Problem 4 in this document. 
I just had to ask it before 2013 ends :)

Among the first 2014 numbers of this kind, two must have the same remainder mod $2013$. 
