Isn't this an old problem? The key idea is to note that the number must also be divisible by $9$ and since $gcd(9,11111)=1$, hence $n \equiv 0 \pmod{99999}$. Let $n=\overline{abcdefghij}$. Then $n \equiv abcde+fghij \equiv 0 \pmod{999999}$. But $abcde+fghij<2(99999)$. Hence, $abcde+fghij=99999$, giving $f=9-a, g=9-b, h=9-c, i=9-d, j=9-e$. Hence, to construct $n$, we only need to choose $\{a,b,c,d,e\}$ from the 5 sets $\{0,9\},\{1,8\},\{2,7\},\{3,6\},\{4,5\}$ keeping in mind that $a \ne 0$. To do this, we first permute in any order the 5 sets and then choose one of the 2 elements. Then we subtract the cases in which $a=0$. Hence, the number of such $n$'s is $2^5 \cdot 5!-2^4 \cdot 4!=(32)(120)-(16)(24)=(16)(24)(9)=\boxed{3456}$.

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