jmc

From Pascal's identity $\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)+\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)=\left(\genfrac{}{}{0ex}{}{n}{k}\right)$.

Therefore, we have $\left(\genfrac{}{}{0ex}{}{20}{11}\right)+\left(\genfrac{}{}{0ex}{}{20}{10}\right)=\left(\genfrac{}{}{0ex}{}{21}{11}\right)$, so $n=11$.

We know that $\left(\genfrac{}{}{0ex}{}{21}{11}\right)=\left(\genfrac{}{}{0ex}{}{21}{21-11}\right)=\left(\genfrac{}{}{0ex}{}{21}{10}\right)$.

We use Pascal's identity again to get $\left(\genfrac{}{}{0ex}{}{20}{9}\right)+\left(\genfrac{}{}{0ex}{}{20}{10}\right)=\left(\genfrac{}{}{0ex}{}{21}{10}\right)$, so $n=9$.

There are two values for $n$, $9$ and $11$, so the sum is $9+11=\boxed{20}$.