Given a positive integer $n, n \geqslant 2$. Try to find all positive integers $m$, such that for any set of positive real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfying the condition:
$$
a_{1} \cdot a_{2} \cdot \cdots \cdot a_{n}=1
$$

the following inequality holds:
$$
a_{1}^{m}+a_{2}^{m}+\cdots+a_{n}^{m} \geqslant \frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}
$$
50 First, find the necessary condition for $m$. Assuming the inequality in the problem holds for any set of positive real numbers $a_{1}, a_{2}, \cdots, a_{n}$ that satisfy $a_{1} \cdot a_{2} \cdot \cdots \cdot a_{n}=1$, then let $a_{1}=a_{2}=\cdots=a_{n-1}=x>0$, $a_{n}=\frac{1}{x^{n-1}}$, and substitute into the inequality to get
$$
(n-1) \cdot x^{m}+\frac{1}{x^{(n-1) m}} \geqslant \frac{n-1}{x}+x^{n-1}
$$

In (1), let $x \rightarrow+\infty$, then $\frac{1}{x^{(n-1) m}}$ and $\frac{n-1}{x}$ both tend to 0. To make (1) hold for every positive number $x$, it is necessary that $m \geqslant n-1(n \geqslant 2)$.
On the other hand, let $m$ be a positive integer not less than $n-1(n \geqslant 2)$, then by the arithmetic and geometric mean inequality we get
$$
\begin{aligned}
& (n-1) \cdot\left(a_{1}^{m}+a_{2}^{m}+\cdots+a_{n}^{m}\right)+n \cdot(m-n+1) \\
= & (a_{1}^{m}+a_{2}^{m}+\cdots+a_{n-1}^{m}+\underbrace{1+\cdots+1}_{m-n+1 \uparrow 1})+ \\
& (a_{1}^{m}+a_{2}^{m}+\cdots+a_{n-2}^{m}+a_{n}^{m}+\underbrace{1+\cdots+1}_{m+n+1 \uparrow 1}) \\
& +\cdots+(a_{2}^{m}+a_{3}^{m}+\cdots+a_{n}^{m}+\underbrace{1+\cdots+1}_{m+n+1}) \\
\geqslant & m \cdot \sqrt[m]{\left(a_{1} \cdot a_{2} \cdot a_{3} \cdots a_{n-1}\right)^{m} \cdot 1^{m-n+1}}+\cdots+ \\
& m \cdot \sqrt[m]{\left(a_{2} \cdot a_{3} \cdots \cdots \cdot a_{n}\right)^{m} \cdot 1^{m-n+1}} \\
= & m\left(a_{1} \cdot a_{2} \cdot \cdots \cdot a_{n-1}+\cdots+a_{2} \cdot a_{3} \cdots \cdots \cdot a_{n}\right) \\
= & m\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}\right),
\end{aligned}
$$

Therefore,
$$
a_{1}^{m}+a_{2}^{m}+\cdots+a_{n}^{m} \geqslant \frac{m}{n-1} \cdot\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}\right)-\frac{n}{n-1} \cdot(m-n+1),
$$

Thus, it is sufficient to prove:
$$
\frac{m}{n-1} \cdot\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}\right)-\frac{n}{n-1} \cdot(m-n+1) \geqslant \frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}},
$$

This is equivalent to
$$
(m-n+1) \cdot\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}-n\right) \geqslant 0 .
$$

Clearly, $m-n+1 \geqslant 0$, and $\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}} \geqslant n \cdot \sqrt[n]{\frac{1}{a_{1}} \cdot \frac{1}{a_{2}} \cdots \cdot \frac{1}{a_{n}}}=n$, so inequality (2) holds. Therefore, the positive integer $m$ is all positive integers greater than or equal to $n-1$.