Question 23 Let $f(n)=\left\{\begin{array}{ll}n-12 & n>2000 ; \\ f[f(n+14)] & n \leqslant 2000\end{array}(n \in \mathbf{N})\right.$. Try to find all the fixed points of $f(n)$.
It is known that when $n \leqslant 2000$ and $n+14>2000$, 
$$
f(n)=f(f(n+14))=f(n+2) .
$$

Therefore, $\quad f(1959)=f(1991)=\cdots=f(2001)=1989$,
$$
f(1990)=f(1992)=\cdots=f(2002)=1990 \text {. }
$$

Thus, $n=1989$ or 1990 are the fixed points of $f(n)$. We will now prove that these are the only fixed points of $f(n)$.
Clearly, when $n>2000$, $f(n)$ has no fixed points.
When $1 \leqslant n<1986$, we can find $k \in \mathbf{N}$ such that,
$$
1986<n+14 k \leqslant 2000 .
$$

Thus, at this time, $\quad f(n)=\underbrace{f(\cdots f(f}_{k+1 \uparrow f}(n+14 k)) \cdots)$.
Since $f(1987)=1989, f(1988)=1990$, from (6), (7), (8), we know that $f(n+14 k)$ only takes the values 1989 or 1990. Therefore, by (9), $f(n)$ also only takes the values 1989 or 1990, meaning that when $1 \leqslant n<1986$, there are no fixed points, and $f(1986)=1990$.