19. (2006 USA Mathematical Olympiad) Find all positive integers $n$ for which there exist a positive integer $k(k \geqslant 2)$ and positive rational numbers $a_{1}, a_{2}, \cdots, a_{k}$, such that: $a_{i}+a_{2}+\cdots+a_{k}=a_{1} a_{2} \cdots a_{k}=n$.
19. The answer is $n=4$ or $n \geqslant 6$.

By the AM-GM inequality, we have $n^{\frac{1}{k}}=\sqrt[k]{a_{1} a_{2} \cdots a_{k}} \leqslant \frac{a_{1}+\cdots+a_{k}}{k}=\frac{n}{k}$.
Therefore, $n \geqslant k^{\frac{k}{k-1}}=k^{1-\frac{1}{k-T}}$.
If $k \geqslant 3$, from the following expressions:
For $k=3$, $n \geqslant 3 \sqrt{3}=\sqrt{27}>5$; for $k=4$, $n \geqslant 4 \sqrt[3]{4}=\sqrt[3]{256}>5$; for $k \geqslant 5$, $n \geqslant 5 \times 5^{\frac{1}{4}}>5$.
For $k=2$, 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 the square of a rational number. In this case, for $n \in\{1,2,3,5\}$, it does not hold, so $n \in\{1,2,3,5\}$.
On the other hand, for $n=4$ and $n \geqslant 6$, we have the following examples:
(1) 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) 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 $a_{1}+\cdots+a_{k}=\left(k+\frac{3}{2}\right)+\frac{1}{2}+4+1 \cdot(k-3)=2 k+3=n ; a_{1} \cdots a_{k}=\left(k+\frac{3}{2}\right) \cdot \frac{1}{2} \cdot 4=2 k+3=n$.
(3) 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 $\{4,6,7, \cdots\}$.