jmc

By Pascal's Identity, we have $\left(\genfrac{}{}{0ex}{}{23}{4}\right)+\left(\genfrac{}{}{0ex}{}{23}{5}\right)=\left(\genfrac{}{}{0ex}{}{24}{5}\right)$. However, we also have $\left(\genfrac{}{}{0ex}{}{24}{5}\right)=\left(\genfrac{}{}{0ex}{}{24}{24-5}\right)=\left(\genfrac{}{}{0ex}{}{24}{19}\right)$. There are no other values of $k$ such that $\left(\genfrac{}{}{0ex}{}{24}{5}\right)=\left(\genfrac{}{}{0ex}{}{24}{k}\right)$, so the sum of all integers that satisfy the problem is $5+19=\boxed{24}$.

Challenge: Is it a coincidence that the answer is 24?