smc

When a $5$-digit number is divided by $100$, the first $3$ digits become the quotient, $q$, and the last $2$ digits become the remainder, $r$. Therefore, $q$ can be any integer from $100$ to $999$ inclusive, and $r$ can be any integer from $0$ to $99$ inclusive. For each of the $9\cdot 10\cdot 10=900$ possible values of $q$, there are at least $\lfloor \frac{100}{11}\rfloor =9$ possible values of $r$ such that $q+r\equiv 0(\mathrm{mod}11)$. Since there is $1$ "extra" possible value of $r$ that is congruent to $0(\mathrm{mod}11)$, each of the $\lfloor \frac{900}{11}\rfloor =81$ values of $q$ that are congruent to $0(\mathrm{mod}11)$ have $1$ more possible value of $r$ such that $q+r\equiv 0(\mathrm{mod}11)$. Therefore, the number of possible values of $n$ such that $q+r\equiv 0(\mathrm{mod}11)$ is $900\cdot 9+81\cdot 1=8181\Rightarrow \boxed{8181}$.