Find all natural values of $k$ for which the system
$$\begin{cases}x_1+x_2+\ldots+x_k=9\\\frac1{x_1}+\frac1{x_2}+\ldots+\frac1{x_k}=1\end{cases}$$
has solutions in positive numbers. Find these solutions.

[i]I. Dimovski[/i]
1. **Applying the Cauchy-Schwarz Inequality:**
   We start by applying the Cauchy-Schwarz inequality to the given system of equations:
   \[
   \left(\sum_{i=1}^k \left(\sqrt{x_i}\right)^2\right)\left(\sum_{i=1}^k \left(\frac{1}{\sqrt{x_i}}\right)^2\right) \geq \left(\sum_{i=1}^k \sqrt{x_i} \cdot \frac{1}{\sqrt{x_i}}\right)^2.
   \]
   Simplifying, we get:
   \[
   \left(\sum_{i=1}^k x_i\right)\left(\sum_{i=1}^k \frac{1}{x_i}\right) \geq \left(\sum_{i=1}^k 1\right)^2.
   \]
   Given the system of equations:
   \[
   \sum_{i=1}^k x_i = 9 \quad \text{and} \quad \sum_{i=1}^k \frac{1}{x_i} = 1,
   \]
   we substitute these into the inequality:
   \[
   9 \cdot 1 \geq k^2 \implies 9 \geq k^2 \implies k \leq 3.
   \]
   Since \(k\) is a natural number, we have:
   \[
   k = 1, 2, 3.
   \]

2. **Checking \(k = 1\):**
   For \(k = 1\), the system becomes:
   \[
   x_1 = 9 \quad \text{and} \quad \frac{1}{x_1} = 1.
   \]
   This implies:
   \[
   x_1 = 1,
   \]
   which contradicts \(x_1 = 9\). Therefore, \(k = 1\) does not yield a valid solution.

3. **Checking \(k = 2\):**
   For \(k = 2\), the system becomes:
   \[
   x_1 + x_2 = 9 \quad \text{and} \quad \frac{1}{x_1} + \frac{1}{x_2} = 1.
   \]
   The second equation can be rewritten as:
   \[
   \frac{x_1 + x_2}{x_1 x_2} = 1 \implies \frac{9}{x_1 x_2} = 1 \implies x_1 x_2 = 9.
   \]
   Thus, \(x_1\) and \(x_2\) are the roots of the quadratic equation:
   \[
   t^2 - 9t + 9 = 0.
   \]
   Solving this using the quadratic formula:
   \[
   t = \frac{9 \pm \sqrt{81 - 36}}{2} = \frac{9 \pm \sqrt{45}}{2} = \frac{9 \pm 3\sqrt{5}}{2}.
   \]
   Therefore, the solutions are:
   \[
   x_1 = \frac{9 + 3\sqrt{5}}{2}, \quad x_2 = \frac{9 - 3\sqrt{5}}{2},
   \]
   and vice versa.

4. **Checking \(k = 3\):**
   For \(k = 3\), the system becomes:
   \[
   x_1 + x_2 + x_3 = 9 \quad \text{and} \quad \frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3} = 1.
   \]
   By the equality case of the Cauchy-Schwarz inequality, we have:
   \[
   x_1 = x_2 = x_3.
   \]
   Let \(x_1 = x_2 = x_3 = x\). Then:
   \[
   3x = 9 \implies x = 3.
   \]
   Therefore, the solution is:
   \[
   x_1 = x_2 = x_3 = 3.
   \]

Conclusion:
The valid values of \(k\) are \(2\) and \(3\). The corresponding solutions are:
\[
\boxed{k \in \{2, 3\}}
\]
\[
\boxed{\left(x_1, x_2\right) = \left(\frac{9 + 3\sqrt{5}}{2}, \frac{9 - 3\sqrt{5}}{2}\right), \left(\frac{9 - 3\sqrt{5}}{2}, \frac{9 + 3\sqrt{5}}{2}\right)}
\]
\[
\boxed{\left(x_1, x_2, x_3\right) = \left(3, 3, 3\right)}
\]