1. Let the function $f: \mathbf{Z}_{+} \rightarrow \mathbf{Z}_{+}$ satisfy
$$
\underbrace{f(f(\cdots f(n)}_{f(n) \cdot \text{times}} \cdots))=\frac{n^{2}}{f(f(n))}
$$

for any positive integer $n$. Find all possible values of $f(1000)$.
1. If $f(a)=f(b)$, then
$$
\begin{array}{l}
a^{2}=f^{f(a)}(a) f(f(a)) \\
=f^{f(b)}(b) f(f(b))=b^{2},
\end{array}
$$

which means $a=b$, hence $f$ is injective.
Next, we prove that for any odd natural number $n$, we have
$$
f(n)=n \text {. }
$$

We use induction on $n$. Assume $f(k)=k$ for all $k \in S=\{1,3, \cdots, n-2\}$ (when $n=1$, we define $S=\varnothing$).
From the given condition, we have
$$
f^{f(n)}(n) f(f(n))=n^{2} \text {. }
$$

Since $f$ is injective, $f^{f(n)}(n)$ and $f(f(n))$ do not belong to $S$, thus, $f(f(n))=n$.

Let $y=f(n)$, then $f(y)=n$, and we have (note that $n$ is odd)
$$
y^{2}=f^{f(y)}(y) f(f(y))=f^{n}(y) y=n y,
$$

which implies $n=y=f(n)$.
The induction is complete.
In summary, if $n$ is a positive even number, then $f(n)$ must be a positive even number, in particular, $f(1000)$ must be a positive even number.
For any fixed positive even number $m$, if we let
$$
\begin{array}{l}
f(1000)=m, f(m)=1000, \\
f(n)=n(n \neq m, 1000),
\end{array}
$$

then it is easy to see that $f$ satisfies the conditions of the problem.
Therefore, all possible values of $f(1000)$ are all positive even numbers.