To solve the given system of equations, we will analyze the function and the behavior of the tangent and cotangent functions.

Given the system of equations:
$$\begin{cases} \tan (x_1) + 3 \cot (x_1) &= 2 \tan (x_2) \\ \tan (x_2) + 3 \cot (x_2) &= 2 \tan (x_3) \\ & \vdots \\ \tan (x_n) + 3 \cot (x_n) &= 2 \tan (x_1) \end{cases}$$

1. Rewrite the equations in terms of a single variable:
Let $y_k = \tan(x_k)$. Then the system becomes:
$$\begin{cases} y_1 + \frac{3}{y_1} &= 2 y_2 \\ y_2 + \frac{3}{y_2} &= 2 y_3 \\ & \vdots \\ y_n + \frac{3}{y_n} &= 2 y_1 \end{cases}$$

2. **Analyze the function $f(z) = \frac{z}{2} + \frac{3}{2z}$:**
We need to find the fixed points of this function. A fixed point $z$ satisfies:
$$z = \frac{z}{2} + \frac{3}{2z}$$
Multiplying both sides by $2z$ to clear the fraction:
$$2z^2 = z^2 + 3 \implies z^2 = 3 \implies z = \pm \sqrt{3}$$

3. **Behavior of the function $f(z)$:**
- For $z > 0$, $f(z)$ has a unique positive fixed point at $z = \sqrt{3}$.
- For $z < 0$, $f(z)$ has a unique negative fixed point at $z = -\sqrt{3}$.

4. Convergence to fixed points:
- If $y_1 \neq \sqrt{3}$, then $y_1 > \sqrt{3}$ and $y_n > \sqrt{3}$ implies $\sqrt{3} < y_{n+1} < y_n$, showing that all positive seeds have orbits converging to $\sqrt{3}$.
- By symmetry, all negative seeds have orbits converging to $-\sqrt{3}$.

5. Conclusion for the system:
Since all $y_k$ converge to either $\sqrt{3}$ or $-\sqrt{3}$, we have:
$$\tan(x_1) = \tan(x_2) = \cdots = \tan(x_n) = \pm \sqrt{3}$$
Therefore, the solutions for $x_k$ are:
$$(x_1, x_2, \ldots, x_n) \in \pi \mathbb{Z}^n \pm \left( \frac{\pi}{3}, \frac{\pi}{3}, \ldots, \frac{\pi}{3} \right)$$

The final answer is $\boxed{ (x_1, x_2, \ldots, x_n) \in \pi \mathbb{Z}^n \pm \left( \frac{\pi}{3}, \frac{\pi}{3}, \ldots, \frac{\pi}{3} \right) }$.