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.
For $n=1$, $\frac{1}{x_{1}^{2}}=\frac{2}{x_{2}^{2}} \Rightarrow \frac{x_{2}^{2}}{x_{1}^{2}}=2$, clearly there are no positive integer solutions; for $n=2$, $\frac{1}{x_{1}^{2}}+\frac{1}{x_{2}^{2}}=\frac{3}{x_{3}^{2}}$ $\Rightarrow\left(x_{2} x_{3}\right)^{2}+\left(x_{1} x_{3}\right)^{2}=3\left(x_{1} x_{2}\right)^{2}$, below we prove that $a^{2}+b^{2}=3 c^{2}$ has no positive integer solutions.
If not, assume $(a, b, c)=1$. From $a^{2} \equiv 0,1(\bmod 3)$, we know that if one of $a, b$ is not a multiple of 3, the equation cannot hold, i.e., $3|a, 3| b$. Let $a=3 a_{1}, b=3 b_{1}$, substitute into the equation to get
$3 a_{1}^{2}+3 b_{1}^{2}=c^{2} \Rightarrow 3 \mid c$, thus $a, b, c$ are all multiples of 3, which contradicts $(a, b, c)=1$.
Therefore, the equation has no positive integer solutions when $n=2$.
For $n=3$, construct $\frac{3^{2}}{3^{2} \cdot 4^{2} \cdot 5^{2}}+\frac{4^{2}}{3^{2} \cdot 4^{2} \cdot 5^{2}}=\frac{5^{2}}{3^{2} \cdot 4^{2} \cdot 5^{2}} \Rightarrow \frac{1}{20^{2}}+\frac{1}{15^{2}}=\frac{1}{12^{2}}$
$\Rightarrow \frac{1}{20^{2}}\left(\frac{1}{20^{2}}+\frac{1}{15^{2}}\right)+\frac{1}{15^{2} \cdot 12^{2}}=\frac{1}{12^{2}}\left(\frac{1}{20^{2}}+\frac{1}{15^{2}}\right)=\frac{1}{12^{2} \cdot 12^{2}}$
$\Rightarrow \frac{1}{400^{2}}+\frac{1}{300^{2}}+\frac{1}{180^{2}}=\frac{1}{144^{2}}=\frac{4}{288^{2}}$, thus for $n=3$, the equation has positive integer solutions.
Therefore, for $n \geqslant 4$, take $x_{1}=400, x_{2}=300, x_{3}=180, x_{4}=x_{5}=\cdots=x_{n}=x_{n+1}=288$, then $\frac{1}{400^{2}}+\frac{1}{300^{2}}+\frac{1}{180^{2}}+\underbrace{\frac{1}{288^{2}}+\frac{1}{288^{2}}+\cdots+\frac{1}{288^{2}}}_{n-3 \uparrow}=\frac{4}{288^{2}}+\frac{n-3}{288^{2}}=\frac{n+1}{288^{2}}$.
In summary, the set of positive integers satisfying the condition is $\left\{n \mid n \geqslant 3, n \in \mathbf{N}_{+}\right\}$.