The 22nd question: Find all positive integers $n \geq 2$, such that for any positive real numbers $a_{1}, a_{2}, \ldots, a_{n}$ satisfying $\prod_{i=1}^{n} a_{i}=1$, the inequality $\sum_{i=1}^{n} \frac{a_{i}}{a_{i+1}}+\sum_{i=1}^{n} \frac{a_{i+1}}{a_{i}} \geq 2 \sum_{i=1}^{n} a_{i}$ always holds, where $a_{n+1}=a_{n}$.
The 22nd question:
Solution: Let $a_{i}=\frac{x_{i}^{n}}{\prod_{i=1}^{n} x_{i}}(1 \leq i \leq n)$, then we need to find the largest positive integer $n$, such that for any positive real numbers $x_{1}, x_{2}, \ldots, x_{n}$, we have
$$
\sum_{i=1}^{n} \frac{x_{i}^{n}}{x_{i+1}^{n}}+\sum_{i=1}^{n} \frac{x_{i+1} 1^{n}}{x_{i}^{n}} \geq 2 \frac{\sum_{i=1}^{n} x_{i}^{n}}{\prod_{i=1}^{n} x_{i}}
$$
(1) When $n \geq 5$, we take $x_{1}=x, x_{2}=x_{n}=\sqrt{x}, x_{3}=x_{4}=\ldots=x_{n-1}=1$, then
$$
\begin{array}{l}
\sum_{i=1}^{n} \frac{x_{i}^{n}}{x_{i+1}^{n}}+\sum_{i=1}^{n} \frac{x_{i+1} 1_{1}^{n}}{x_{i}^{n}}=2\left(x^{\frac{n}{2}}+x^{\frac{n}{2}}+(n-4)+\frac{1}{x^{\frac{n}{2}}}+\frac{1}{x^{\frac{n}{2}}}\right) \\
2 \frac{\sum_{i=1}^{n} x_{i}^{n}}{\prod_{i=1}^{n} x_{i}}=2 \cdot \frac{x^{n}+2 x^{\frac{n}{2}}+(n-2)}{x^{2}}
\end{array}
$$

Since $n-2>\frac{n}{2}$, when $x \rightarrow+\infty$, we must have $2\left(x^{\frac{n}{2}}+x^{\frac{n}{2}}+(n-4)+\frac{1}{x^{\frac{n}{2}}}+\frac{1}{x^{\frac{n}{2}}}\right)<2 \cdot \frac{x^{n}+2 x^{\frac{n}{2}}+(n-2)}{x^{2}}$. Therefore, the proposition does not hold.
(2) Now prove that when $n=2, 3, 4$, the conclusion holds.
(1) When $n=2$, we need to prove:
$$
\frac{a^{2}}{b^{2}}+\frac{b^{2}}{a^{2}}+\frac{b^{2}}{a^{2}}+\frac{a^{2}}{b^{2}} \geq 2 \frac{a^{2}+b^{2}}{a b} \Leftrightarrow a^{4}+b^{4} \geq a b\left(a^{2}+b^{2}\right)
$$

According to the rearrangement inequality, the above inequality clearly holds.
(2) When $n=3$, we need to prove:
$$
\begin{array}{l}
\frac{a^{3}}{b^{3}}+\frac{b^{3}}{c^{3}}+\frac{c^{3}}{a^{3}}+\frac{b^{3}}{a^{3}}+\frac{c^{3}}{b^{3}}+\frac{a^{3}}{c^{3}} \geq 2 \frac{a^{3}+b^{3}+c^{3}}{a b c} \\
\Leftrightarrow \sum a^{3}\left(\frac{1}{b^{3}}+\frac{1}{c^{3}}\right) \geq 2 \frac{a^{3}+b^{3}+c^{3}}{a b c}
\end{array}
$$

In fact, according to the AM-GM inequality, we know:
$$
\sum a^{3}\left(\frac{1}{b^{3}}+\frac{1}{c^{3}}\right)=\sum a^{3}\left(\frac{1}{b^{3}}+\frac{1}{c^{3}}+\frac{1}{a^{3}}\right)-3 \geq 3 \sum \frac{a^{3}}{a b c}-3 \geq 2 \frac{a^{3}+b^{3}+c^{3}}{a b c}
$$

The proposition holds!
(3) When $n=4$, we need to prove:
$$
\frac{a^{4}}{b^{4}}+\frac{b^{4}}{c^{4}}+\frac{c^{4}}{d^{4}}+\frac{d^{4}}{a^{4}}+\frac{b^{4}}{a^{4}}+\frac{c^{4}}{b^{4}}+\frac{d^{4}}{c^{4}}+\frac{a^{4}}{d^{4}} \geq 2 \frac{a^{4}+b^{4}+c^{4}+d^{4}}{a b c d}
$$

In fact, according to the AM-GM inequality, we know:
$$
\text { LHS }=\frac{\left(\sum_{\text {cyc }}\left(\frac{a^{4}}{b^{4}}+\frac{a^{4}}{b^{4}}+\frac{b^{4}}{c^{4}}+\frac{a^{4}}{d^{4}}\right)\right)+\left(\sum_{\text {cyc }}\left(\frac{b^{4}}{a^{4}}+\frac{b^{4}}{a^{4}}+\frac{a^{4}}{d^{4}}+\frac{b^{4}}{c^{4}}\right)\right)}{4} \geq \sum_{\text {cyc }} \frac{a^{3}}{b c d}+\sum_{\text {cyc }} \frac{b^{3}}{c d a}=2 \frac{a^{4}+b^{4}+c^{4}+d^{4}}{a b c d}
$$

The proposition holds!
In summary, the positive integers $n=2, 3, 4$.