jmc

Since $\frac{1}{13}=13^{-1}$, we can express ${\left(\frac{1}{13}\right)}^{n-24}$ as $13^{-n+24}$. We have that $13^{3n}={\left(\frac{1}{13}\right)}^{n-24}=13^{-n+24}$, so setting the exponents equal we find that $3n=-n+24$, or $n=\frac{24}{4}=\boxed{6}$.