4. Find the function $f: \mathbf{Z}_{+} \rightarrow \mathbf{Z}_{+}$, such that for all $m, n \in \mathbf{Z}_{+}$, we have
$$
(n!+f(m)!) \mid(f(n)!+f(m!)) \text {. }
$$
4. Taking $m=n=1$, substituting into equation (1) we get
$$
\begin{array}{l}
(1+f(1)!) \mid(f(1)!+f(1)) \\
\Rightarrow(1+f(1)!) \mid(f(1)-1) .
\end{array}
$$
$$
\begin{array}{l}
\text { Since }|f(1)-1|<f(1)!+1 \\
\Rightarrow f(1)-1=0 \\
\Rightarrow f(1)=1 .
\end{array}
$$

When $m=1$, equation (1) becomes
$$
\begin{array}{l}
(n!+1) \mid(f(n)!+1) \\
\Rightarrow n!\leqslant f(n)! \\
\Rightarrow n \leqslant f(n) .
\end{array}
$$

For any prime $p$, taking $(m, n)=(1, p-1)$, by Wilson's theorem we get
$$
\begin{array}{l}
p \mid((p-1)!+1) \\
\Rightarrow p \mid(f(p-1)!+1) \\
\Rightarrow f(p-1)<p .
\end{array}
$$

Since $f(p-1) \geqslant p-1$, thus,
$$
f(p-1)=p-1 \text {. }
$$

Next, fix a positive integer $m$. For any prime $p$, in equation (1) let $n=p-1$ we get
$$
\begin{array}{l}
((p-1)!+f(m)!) \mid((p-1)!+f(m!)) \\
\Rightarrow((p-1)!+f(m)!) \mid(f(m!)-f(m)!) .
\end{array}
$$

Then for all $m \in \mathbf{Z}_{+}$, we have
$$
f(m!)=f(m)!\text {. }
$$

Therefore, equation (1) can be rewritten as
$$
\begin{array}{l}
(n!+f(m)!) \mid(f(n)!+f(m)!) \\
\Rightarrow(n!+f(m)!) \mid(f(n)!-n!) .
\end{array}
$$

Thus, for all $n \in \mathbf{Z}_{+}$, and all sufficiently large positive integers $m$ we get
$$
f(n)!=n!\Rightarrow f(n)=n .
$$