Find the number $n$, such that there exist a natural number $k \geqslant 2$ and positive rational numbers $a_{1}, a_{2}, \cdots$, $a_{k}$, satisfying
$$
a_{1}+a_{2}+\cdots+a_{k}=a_{1} a_{2} \cdots a_{k}=n .
$$
Parse $n=a_{1}+a_{2}+\cdots+a_{k} \geqslant k \sqrt[k]{a_{1} a_{2} \cdots a_{k}}=k \sqrt[k]{n}$, so $n^{1-\frac{1}{k}} \geqslant k, n \geqslant$ $k^{1+\frac{1}{k-1}}$, we get $3^{1+\frac{1}{2}}>5,4^{1+\frac{1}{3}}>5,2^{1+\frac{1}{1}}=4$. Therefore, when $k>2$, $n>5$.

When $k=2$, $a_{1}+a_{2}=a_{1} a_{2}=n$ leads to $a_{1} 、 a_{2}$ being the rational roots of the quadratic equation $x^{2}-n x+n=0$. Therefore, they are also positive integer roots. At this time, $\sqrt{n^{2}-4 n}$ should be an integer, $n=4$. So the values of $n$ all belong to the set $\{4,6,7,8, \cdots\}$.
When $n=$ even $2 m(m \geqslant 2)$, take $k=m,\left(a_{1}, a_{2}, \cdots, a_{k}\right)=(k, 2,1, \cdots, 1)$.
When $n=$ odd $2 m+3 \geqslant 9$, take $k=m,\left(a_{1}, a_{2}, \cdots, a_{k}\right)=\left(a_{1}, a_{2}, a_{3}, 1,1, \cdots, 1\right)$. The undetermined rational numbers $a_{1}, a_{2}, a_{3}$ satisfy
$$
\left\{\begin{array}{l}
a_{1}+a_{2}+a_{3}=k+6 \\
a_{1} a_{2} a_{3}=2 k+3
\end{array}\right.
$$
The coefficient of $k$ in (2) is twice that in (1). We choose $a_{2} a_{3}=2, a_{1}=\frac{2 k+3}{2}$ in (2), then substituting into (1) we get, $a_{2}+a_{3}=\frac{9}{2}$. Thus $a_{2}=4, a_{3}=\frac{1}{2}$, that is
$$
\left(a_{1}, a_{2}, \cdots, a_{k}\right)=\left(\frac{2 k+3}{2}, 4, \frac{1}{2}, 1,1, \cdots, 1\right) .
$$
When $n=7$, take $k=3,\left\{\begin{array}{l}a_{1}+a_{2}+a_{3}=7, \\ a_{1} a_{2} a_{3}=7 .\end{array}\right.$ Let $a_{1}=\frac{a}{d}, a_{2}=\frac{b}{d}, a_{3}=\frac{c}{d}, a 、 b$ 、
$c$ be positive integers, then
$$
\begin{array}{c}
\left\{\begin{array}{l}
a+b+c=7 d, \\
a b c-7 d^{3} .
\end{array}\right. \\
\left\{\begin{array}{l}
a+b=7(d-1), \\
a b=d^{3} .
\end{array}\right.
\end{array}
$$

We can take $a 、 b$ to be cubes, the simplest being $a=2^{3}=8, b=3^{3}=27$. At this time, $d=6$ satisfies (5) and (6). So $\left(a_{1}, a_{2}, a_{3}\right)=\left(\frac{4}{3}, \frac{9}{2}, \frac{7}{6}\right)$.

Comment: We do not need to find all positive integer solutions to the indeterminate equation systems (3) and (4) or (5) and (6). We only need to find (read out) one solution. Therefore, we should try to take those values that are possibly larger and can simplify the problem (such as $c=7$, such as $\left.a=2^{3}, b=3^{3}\right)$).