Open Contests

Note that the minimal value of a $k$-digit number is $10^{k-1}$ and the maximal value of the cross-sum multiplied by $2013$ is $9k\cdot 2013$. Since $9\cdot 7\cdot 2013=126819<1000000$ we can consider only numbers with up to $6$ digits. Since then the cross-sum is at most $54$, it is enough to consider numbers in the form $n\cdot 2013$ with $1\le n\le 54$.

Since $2013$ is divisible by $3$, $n\cdot 2013$ and its cross-sum are divisible by $3$. Since the cross-sum must be equal to $n$, $n\cdot 2013$ is divisible by $9$. But then its cross-sum and hence also $n$ is divisible by $9$. It remains to consider the cases $n=9,18,\dots ,54$ which can be checked by hand and see that only $n=18$ satisfies the conditions.

Answer: $36234$.