By Cauchy - Schwarz, $$\left(\sum_{i=1}^k \left(\sqrt{x_i}\right)^2\right)\left(\sum_{i=1}^k \left(\frac{1}{\sqrt{x_i}}\right)^2\right) \geq \left(\sum_{i=1}^k \sqrt{x_i} \cdot \frac{1}{\sqrt{x_i}}\right)^2.$$This becomes $$\left(\sum_{i=1}^k x_i\right)\left(\sum_{i=1}^k \frac{1}{x_i}\right) \geq \left(\sum_{i=1}^k 1\right)^2.$$Plugging in yields $$9 \cdot 1 \geq k^2 \implies |k| \leq 3 \implies -3 \leq k \leq 3.$$Since $k$ is a natural number, we have $k = 1, 2, 3.$ It remains to determine whether or not all three values of $k$ yield valid solutions $\left(x_1, x_2, x_3, \hdots, x_k\right)$ to the original system, and the solutions themselves. Clearly $k=1$ is not valid as $x_1 = 9$ and $\tfrac{1}{x_1}=1$ may not simultaneously hold; next, $k=2$ yields $x_1+x_2=9$ and $\tfrac{1}{x_1}+\tfrac{1}{x_2}=\tfrac{x_1+x_2}{x_1x_2}=1,$ which leads to $x_1x_2=9.$ Now $x_1$ and $x_2$ are the roots of the quadratic $x^2-9x+9$ in some order; applying the quadratic formula, we have $x_1, x_2 = \tfrac{9 \pm 9\sqrt{5}}{2}.$ Finally, for $k=3,$ we must have $\tfrac{\sqrt{x_1}}{\tfrac{1}{\sqrt{x_1}}}=\tfrac{\sqrt{x_2}}{\tfrac{1}{\sqrt{x_2}}}=\tfrac{\sqrt{x_3}}{\tfrac{1}{\sqrt{x_3}}}$ which implies $x_1 = x_2 = x_3$ by the equality case of Cauchy - Schwarz, which leads to $x_1 = x_2 = x_3 = 3$ upon inspection. All in all, $\boxed{k \in \{2, 3\}}$ and $\boxed{\left(x_1, x_2, x_3, \hdots, x_k\right) = \left(\tfrac{9-9\sqrt{5}}{2}, \tfrac{9+9\sqrt{5}}{2}\right), \left(\tfrac{9+9\sqrt{5}}{2}, \tfrac{9-9\sqrt{5}}{2}\right), \left(3, 3, 3\right)}.$