3. Given the function $f(n)=\frac{20^{n}+3^{n}}{n!}, n \in \mathbf{N}$. Then, the value of $n$ that maximizes $f(n)$ is, $n=$
3.19.
$$
\begin{array}{l}
G(n)=f(n+1)-f(n) \\
=\frac{20^{n+1}+3^{n+1}}{(n+1)!}-\frac{20^{n}+3^{n}}{n!} \\
=\frac{(19-n) \times 20^{n}-(n-2) \times 3^{n}}{(n+1)!} \\
=\frac{3^{n}}{(n+1)!}\left[(19-n) \times\left(\frac{20}{3}\right)^{n}-(n-2)\right] .
\end{array}
$$

It is easy to see that when $n \geqslant 19$, $G(n) < 0$.
Therefore, $f(19)$ is the maximum value.