The 21st question: Find all positive integers $n$ such that for any $a_{1}, a_{2}, \ldots, a_{n}, a_{i} \in (0,1] (1 \leq i \leq n)$, the inequality $\sum_{i=1}^{n} \frac{\sqrt{1-a_{i}}}{a_{i}} \leq \frac{1}{\prod_{i=1}^{n} a_{i}}$ always holds.
The 21st question,
Answer: First, if the inequality holds for $\mathrm{n}$, then it also holds for $\mathrm{n}-1$.
Indeed, if for any $a_{i} \in(0,1](1 \leq i \leq n)$, we have $\sum_{i=1}^{n} \frac{\sqrt{1-a_{i}}}{a_{i}} \leq \frac{1}{\prod_{i=1}^{n} a_{i}}$. We take $a_{n}=1$, then we get $\sum_{\mathrm{i}=1}^{\mathrm{n}-1} \frac{\sqrt{1-\mathrm{a}_{\mathrm{i}}}}{\mathrm{a}_{i}} \leq \frac{1}{\prod_{i=1}^{\mathrm{n}-1} \mathrm{a}_{\mathrm{i}}}$.
Thus, we only need to find the largest positive integer $n$ such that the inequality always holds.
When $n \geq 5$, we take $a_{1}=a_{2}=a_{3}=a_{4}=a_{5}=\frac{8}{9}, a_{6}=a_{7}=\ldots=a_{n}=1$, then the inequality does not hold.
Below, we prove that when $n=4$, the inequality always holds.
First, we prove a lemma: If $x, y \in[0,1]$, then $x \sqrt{1-y}+y \sqrt{1-x} \leq \sqrt{1-x^{2} y^{2}}$.
Proof of the lemma: In fact, because $(1-x)(1-y) \geq 0 \Rightarrow x+y \leq 1+x y$. Thus, according to the Cauchy-Schwarz inequality, we know:
$$
\begin{array}{l}
x \sqrt{1-y}+y \sqrt{1-x} \leq \sqrt{1-x^{2} y^{2}} \\
=\sqrt{x} \cdot \sqrt{x(1-y)}+\sqrt{y} \cdot \sqrt{y(1-x)} \\
\leq \sqrt{(x+y)[x(1-y)+y(1-x)]} \\
=\sqrt{(x+y)(x+y-2 x y)} \\
\leq \sqrt{(1+x y)[(1+x y)-x y]}=\sqrt{1-x^{2} y^{2}}
\end{array}
$$

The lemma is proved!
Now, let's return to the original problem.
We need to prove: If $a, b, c, d \in(0,1]$, then $\frac{\sqrt{a}}{1-a}+\frac{\sqrt{b}}{1-b}+\frac{\sqrt{c}}{1-c}+\frac{\sqrt{d}}{1-d} \leq \frac{1}{a b c d}$.
In fact, because
$$
\begin{array}{l}
\frac{\sqrt{a}}{1-a}+\frac{\sqrt{b}}{1-b}+\frac{\sqrt{c}}{1-c}+\frac{\sqrt{d}}{1-d} \leq \frac{1}{a b c d} \\
\Leftrightarrow c d(a \sqrt{1-b}+b \sqrt{1-a})+a b(c \sqrt{1-d}+d \sqrt{1-c}) \leq 1
\end{array}
$$

According to the lemma, we know:
$$
\begin{array}{l}
c d(a \sqrt{1-b}+b \sqrt{1-a})+a b(c \sqrt{1-d}+d \sqrt{1-c}) \\
\leq c d \cdot \sqrt{1-a^{2} b^{2}}+a b \cdot \sqrt{1-c^{2} d^{2}} \\
\leq \frac{c^{2} d^{2}+\left(1-a^{2} b^{2}\right)}{2}+\frac{a^{2} b^{2}+\left(1-c^{2} d^{2}\right)}{2}=1
\end{array}
$$

Thus, (*) holds.
In summary, the possible values of $\mathrm{n}$ are $1, 2, 3, 4$.