+ Find all positive integers $n$ such that there exist $k \in \mathbf{N}^{*}, k \geqslant 2$ and positive rational numbers $a_{1}, a_{2}, \cdots, a_{k}$, satisfying
$$
a_{1}+a_{2}+\cdots+a_{k}=a_{1} \cdots a_{k}=n .
$$
The solution is $n=4$ or $n \geqslant 6$.
By the AM-GM inequality, we have
$$
n^{\frac{1}{k}}=\sqrt[k]{a_{1} \cdots a_{k}} \leqslant \frac{a_{1}+\cdots+a_{k}}{k}=\frac{n}{k} .
$$

Therefore, $n \geqslant k^{k-1}=k^{1+\frac{1}{k-1}}$.
If $k \geqslant 3$, we have the following:
$$
\begin{array}{l}
k=3 \text {, } n \geqslant 3 \sqrt{3}=\sqrt{27}>5 ; \\
k=4 \text {, } n \geqslant 4 \sqrt[3]{4}=\sqrt[3]{256}>5 ; \\
k \geqslant 5 \text {, } n \geqslant 5 \times 5^{\frac{1}{k-1}}>5 .
\end{array}
$$

When $k=2$, from $a_{1}+a_{2}=a_{1} a_{2}=n$, we know that $n^{2}-4 n=\left(a_{1}+a_{2}\right)^{2}-$ $4 a_{1} a_{2}=\left(a_{1}-a_{2}\right)^{2}$ is a square of a rational number. This does not hold for $n \in\{1,2,3,5\}$. Therefore, $n \notin\{1,2,3,5\}$.
On the other hand, for $n=4$ and $n \geqslant 6$, we have the following examples.
$1^{\circ}$ When $n$ is even, let $n=2 k(\geqslant 4)$, take $\left(a_{1}, a_{2}, \cdots, a_{k}\right)=(k$, $2,1, \cdots, 1)$, then $a_{1}+\cdots+a_{k}=k+2+1 \cdot(k-2)=2 k=n$; $a_{1} \cdots a_{k}=2 k=n$.
$2^{\circ}$ When $n$ is odd, if $n=2 k+3 \geqslant 9$, then we can take $\left(a_{1}, \cdots, a_{k}\right)$ $=\left(k+\frac{3}{2}, \frac{1}{2}, 4,1, \cdots, 1\right)$, then
$$
\begin{aligned}
& a_{1}+\cdots+a_{k}=\left(k+\frac{3}{2}\right)+\frac{1}{2}+4+1 \cdot(k-3) \\
= & 2 k+3=n ; a_{1} \cdots a_{k}=\left(k+\frac{3}{2}\right) \cdot \frac{1}{2} \cdot 4 \\
= & 2 k+3=n .
\end{aligned}
$$
$3^{\circ}$ When $n=7$, take $\left(a_{1}, a_{2}, a_{3}\right)=\left(\frac{4}{3}, \frac{7}{6}, \frac{9}{2}\right)$.

In summary, the required $n$ is $\{4,6,7, \cdots\}$.