16. Find all positive integers $n$ such that the equation
$$
\frac{1}{x_{1}^{2}}+\frac{1}{x_{2}^{2}}+\cdots+\frac{1}{x_{n}^{2}}=\frac{n+1}{x_{n+1}^{2}}
$$

has positive integer solutions.
16. (1) When $n=1$, the equation becomes
$$
\frac{1}{x_{1}^{2}}=\frac{2}{x_{2}^{2}} \Rightarrow \frac{x_{2}}{x_{1}}=\sqrt{2} \text {, }
$$

obviously there is no positive integer solution.
(2) When $n=2$, the equation becomes
$$
\begin{array}{l}
\frac{1}{x_{1}^{2}}+\frac{1}{x_{2}^{2}}=\frac{3}{x_{3}^{2}} \\
\Rightarrow\left(x_{2} x_{3}\right)^{2}+\left(x_{1} x_{3}\right)^{2}=3\left(x_{1} x_{2}\right)^{2} .
\end{array}
$$

Lemma: $a^{2}+b^{2}=3 c^{2}$ has no positive integer solutions.
Proof: Assume there is a set of positive integer solutions $a, b, c$, and without loss of generality, assume their greatest common divisor is 1.
Since $a, b$ are positive integers, $a^{2} \equiv 0$ or $1(\bmod 3), b^{2} \equiv 0$ or $1(\bmod 3)$. Also, $3 c^{2} \equiv 0(\bmod 3)$, so $a^{2} \equiv 0(\bmod 3)$ and $b^{2} \equiv 0(\bmod 3)$ $\Rightarrow a \equiv 0(\bmod 3)$ and $b \equiv 0(\bmod 3)$.
Thus, $c \equiv 0(\bmod 3)$.
This contradicts the assumption that the greatest common divisor of $a, b, c$ is 1. The lemma is proved.
By the lemma, the original equation has no positive integer solutions.
(3) When $n=3$, the equation becomes
$\frac{1}{x_{1}^{2}}+\frac{1}{x_{2}^{2}}+\frac{1}{x_{3}^{2}}=\frac{4}{x_{4}^{2}}$.
From $3^{2}+4^{2}=5^{2}$, we get
$\frac{3^{2}}{(3 \times 4 \times 5)^{2}}+\frac{4^{2}}{(3 \times 4 \times 5)^{2}}=\frac{5^{2}}{(3 \times 4 \times 5)^{2}}$
$\Rightarrow \frac{1}{20^{2}}+\frac{1}{15^{2}}=\frac{1}{12^{2}}$.
Then $\frac{1}{20^{2} \times 12^{2}}+\frac{1}{15^{2} \times 12^{2}}=\frac{1}{\left(12^{2}\right)^{2}}$
$\Rightarrow \frac{1}{20^{2}}\left(\frac{1}{20^{2}}+\frac{1}{15^{2}}\right)+\frac{1}{15^{2} \times 12^{2}}=\frac{1}{\left(12^{2}\right)^{2}}$
$\Rightarrow \frac{1}{400^{2}}+\frac{1}{300^{2}}+\frac{1}{180^{2}}=\frac{1}{144^{2}}$
$\Rightarrow \frac{1}{400^{2}}+\frac{1}{300^{2}}+\frac{1}{180^{2}}=\frac{4}{288^{2}}$.
This shows that the original equation has positive integer solutions
$x_{1}=400, x_{2}=300, x_{3}=180, x_{4}=288$.
(4) When $n \geqslant 4$,
$$
\frac{1}{x_{1}^{2}}+\frac{1}{x_{2}^{2}}+\cdots+\frac{1}{x_{n}^{2}}=\frac{n+1}{x_{n+1}^{2}}
$$

has positive integer solutions
$$
\begin{array}{l}
x_{1}=400, x_{2}=300, x_{3}=180, \\
x_{4}=x_{5}=\cdots=x_{n}=x_{n+1}=288 .
\end{array}
$$

In summary, when $n=1$ or 2, the original equation has no positive integer solutions; when $n \geqslant 3$, the original equation has positive integer solutions.
Therefore, the required $n$ is $\left\{n \in \mathbf{Z}_{+} \mid n \geqslant 3\right\}$.
(Provided by Guo Min)