Determine the natural numbers $n\ge 2$ for which  exist $x_1,x_2,...,x_n \in R^*$,  such that $$x_1+x_2+...+x_n=\frac{1}{x_1}+\frac{1}{x_2}+...+\frac{1}{x_n}=0$$
To determine the natural numbers \( n \geq 2 \) for which there exist \( x_1, x_2, \ldots, x_n \in \mathbb{R}^* \) such that 
\[ x_1 + x_2 + \cdots + x_n = \frac{1}{x_1} + \frac{1}{x_2} + \cdots + \frac{1}{x_n} = 0, \]
we proceed as follows:

1. **Base Case for \( n = 2 \):**
   - Consider \( n = 2 \). We need to find \( x_1, x_2 \in \mathbb{R}^* \) such that:
     \[ x_1 + x_2 = 0 \quad \text{and} \quad \frac{1}{x_1} + \frac{1}{x_2} = 0. \]
   - From \( x_1 + x_2 = 0 \), we get \( x_2 = -x_1 \).
   - Substituting \( x_2 = -x_1 \) into the second equation:
     \[ \frac{1}{x_1} + \frac{1}{-x_1} = 0 \implies \frac{1}{x_1} - \frac{1}{x_1} = 0, \]
     which is true. Therefore, \( n = 2 \) is a solution.

2. **Inductive Step:**
   - Suppose \( n \) is a solution. We need to show that \( n + 2 \) is also a solution.
   - Given \( x_1, x_2, \ldots, x_n \) such that:
     \[ x_1 + x_2 + \cdots + x_n = 0 \quad \text{and} \quad \frac{1}{x_1} + \frac{1}{x_2} + \cdots + \frac{1}{x_n} = 0, \]
     we can add \( x_{n+1} = 1 \) and \( x_{n+2} = -1 \). Then:
     \[ x_1 + x_2 + \cdots + x_n + x_{n+1} + x_{n+2} = 0 + 1 - 1 = 0, \]
     and
     \[ \frac{1}{x_1} + \frac{1}{x_2} + \cdots + \frac{1}{x_n} + \frac{1}{x_{n+1}} + \frac{1}{x_{n+2}} = 0 + 1 - 1 = 0. \]
   - Therefore, if \( n \) is a solution, then \( n + 2 \) is also a solution.

3. **Checking \( n = 3 \):**
   - Suppose \( n = 3 \). We need to find \( x_1, x_2, x_3 \in \mathbb{R}^* \) such that:
     \[ x_1 + x_2 + x_3 = 0 \quad \text{and} \quad \frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3} = 0. \]
   - From \( x_1 + x_2 + x_3 = 0 \), we have \( x_3 = -(x_1 + x_2) \).
   - Substituting \( x_3 = -(x_1 + x_2) \) into the second equation:
     \[ \frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{-(x_1 + x_2)} = 0. \]
   - This simplifies to:
     \[ \frac{1}{x_1} + \frac{1}{x_2} - \frac{1}{x_1 + x_2} = 0. \]
   - Multiplying through by \( x_1 x_2 (x_1 + x_2) \) gives:
     \[ x_2 (x_1 + x_2) + x_1 (x_1 + x_2) - x_1 x_2 = 0 \implies x_1 x_2 + x_2^2 + x_1^2 + x_1 x_2 - x_1 x_2 = 0 \implies x_1^2 + x_2^2 + x_1 x_2 = 0. \]
   - This equation has no real solutions because the discriminant of the quadratic equation \( x_1^2 + x_1 x_2 + x_2^2 = 0 \) is negative. Therefore, \( n = 3 \) is not a solution.

4. **Checking \( n = 5 \):**
   - Consider \( n = 5 \). We need to find \( x_1, x_2, x_3, x_4, x_5 \in \mathbb{R}^* \) such that:
     \[ x_1 + x_2 + x_3 + x_4 + x_5 = 0 \quad \text{and} \quad \frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3} + \frac{1}{x_4} + \frac{1}{x_5} = 0. \]
   - Select \( x_1 = x_2 = x_3 = -1 \), \( x_4 = \frac{3 + \sqrt{5}}{2} \), and \( x_5 = \frac{3 - \sqrt{5}}{2} \).
   - Then:
     \[ x_1 + x_2 + x_3 + x_4 + x_5 = -1 - 1 - 1 + \frac{3 + \sqrt{5}}{2} + \frac{3 - \sqrt{5}}{2} = -3 + 3 = 0, \]
     and
     \[ \frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3} + \frac{1}{x_4} + \frac{1}{x_5} = -1 - 1 - 1 + \frac{2}{3 + \sqrt{5}} + \frac{2}{3 - \sqrt{5}}. \]
   - Simplifying the last two terms:
     \[ \frac{2}{3 + \sqrt{5}} + \frac{2}{3 - \sqrt{5}} = \frac{2(3 - \sqrt{5}) + 2(3 + \sqrt{5})}{(3 + \sqrt{5})(3 - \sqrt{5})} = \frac{6}{9 - 5} = \frac{6}{4} = \frac{3}{2}. \]
   - Therefore:
     \[ -1 - 1 - 1 + \frac{3}{2} + \frac{3}{2} = -3 + 3 = 0. \]
   - Thus, \( n = 5 \) is a solution.

Conclusion:
- From the base case and the inductive step, we conclude that \( n \) must be even.
- Additionally, we have shown that \( n = 5 \) is also a solution.

The final answer is \( \boxed{ n } \) is even or \( n = 5 \).