6.14. (NPR, 68). Find all values of $n \in \mathbf{N}$ for which there exists a set of positive numbers $x_{1}, \ldots, x_{\mu}$, satisfying the system

$$
\left\{\begin{array}{l}
x_{1}+x_{2}+\ldots+x_{n}=9 \\
\frac{1}{x_{1}}+\frac{1}{x_{2}}+\ldots+\frac{1}{x_{n}}=1
\end{array}\right.
$$

Indicate all such sets for each of the found values of $n$.
6.14. Let $R=\sqrt[n]{x_{1} \ldots x_{n}}$, where $\left(x_{1} ; \ldots ; x_{n}\right)$ is a solution to the system. Applying the theorem of means, we get

$$
1=\frac{1}{x_{1}}+\frac{1}{x_{2}}+\ldots+\frac{1}{x_{n}} \geqslant \frac{n}{R}, \quad 9=x_{1}+x_{2}+\ldots+x_{n} \geqslant n R,
$$

from which $R \geqslant n$ and $R \leqslant 9 / n$. Therefore, $n^{2} \leq 9$ and $n \leqslant 3$. In this case, if $n=3$, then all the written inequalities turn into equalities, which is possible (and indeed holds) only in the case $x_{3}=x_{2}=$ $=x_{3}=R=3$. Thus, the value $n=3$ satisfies the condition of the problem, and for this value of $n$ the system has a unique solution in positive numbers ( $3 ; 3 ; 3$ ). Since for $n=1$ we get an inconsistent system, it remains to consider the case $n=2$. In this case, we have the system

$$
\left\{\begin{array}{l}
x_{1}+x_{2}=9 \\
\frac{1}{x_{1}}+\frac{1}{x_{2}}=1
\end{array}\right.
$$

which, due to the equality

$$
\frac{1}{x_{1}}+\frac{1}{x_{2}}=\frac{x_{1}+x_{2}}{x_{1} x_{2}}
$$

is equivalent to the system

$$
\left\{\begin{aligned}
x_{1}+x_{2} & =9 \\
x_{1} x_{2} & =9
\end{aligned}\right.
$$

By the theorem converse to Vieta's theorem, the values $x_{1}, x_{2}$ form a pair of roots of the polynomial $t^{2}-9 t+9$, which are $t_{1,2}=(9 \pm 3 \sqrt{5}) / 2$. Thus, we have two solutions $\left(t_{1} ; t_{2}\right),\left(t_{2} ; t_{1}\right)$. Since $t_{1}, t_{2}>0$, the value $n=2$ satisfies the condition of the problem.