Solve for $n=0$, $f[f(0)]+f(0)=f(1)+f(0)=f(1)+1=2 \cdot 0+3$, so $f(1)=2$.
Let $n=1$, then $f[f(1)]+f(1)=2 \cdot 1+3=5$, which means $f(2)+2=5$, so $f(2)=3$.
From this, we conjecture that $f(n)=n+1$.
We need to prove this by induction for all $n$ in the set of integers $\mathbf{Z}$.
For $n \geqslant 0$, if $n \leqslant k$, then $f(n)=n+1$, so
$$f[f(k)]+f(k)=2 k+3, f(k+1)+k+1=2 k+3,$$

which means $f(k+1)=2 k+3$. Therefore, for $n \geqslant 0$ and $n \in \mathbf{Z}$, $f(n)=n+1$.
Assume for some $k \leqslant 0$, when $n \geqslant k$, $f(n)=n+1$.
We need to prove that when $n=k-1$, $f(k-1)=k$.
Let $f(k-1)=a \in \mathbf{Z}$, then
$$\begin{array}{l} f(a)=f[f(k-1)]=2(k-1)+3-f(k-1)=2 k+1-a, \\ \text { thus } f(2 k+1-a)=f[f(a)]=2 a+3-(2 k+1-a)=3 a-2 k+2, \\ \begin{aligned} f(3 a-2 k+2) & =f[f(2 k+1)-a]=2(2 k+1-a)+3-f(2 k+1-a) \\ & =6 k+5 a+3 . \end{aligned} \end{array}$$

If $2 k+1-a \geqslant k$, by the induction hypothesis,
$f(2 k+1-a)=2 k+2-a$, which means $3 a-2 k+2=2 k+2-a$, so $a=k$.
If $3 a+2-2 k \geqslant k$, by the induction hypothesis,
$f(3 a+2-2 k)=3 a+3-2 k=6 k-5 a+3$, so $a=k$.
If $\left\{\begin{array}{l}2 k+1-ak+1, \\ a<k-\frac{2}{3},\end{array}\right.$, a contradiction arises.
Thus, $f(k-1)=k$.
In conclusion, for all $n \in \mathbf{Z}$, $f(n)=n+1$.