jmc

By Pascal's Identity, we have $\left(\genfrac{}{}{0ex}{}{8}{3}\right)+\left(\genfrac{}{}{0ex}{}{8}{4}\right)=\left(\genfrac{}{}{0ex}{}{9}{4}\right)$. However, we also have $\left(\genfrac{}{}{0ex}{}{9}{4}\right)=\left(\genfrac{}{}{0ex}{}{9}{9-4}\right)=\left(\genfrac{}{}{0ex}{}{9}{5}\right)$. There are no other values of $n$ such that $\left(\genfrac{}{}{0ex}{}{9}{4}\right)=\left(\genfrac{}{}{0ex}{}{9}{n}\right)$, so the largest possible value of $n$ is $\boxed{5}$.