Find all positive integers $n$ for which there exist real numbers $x_1, x_2,. . . , x_n$ satisfying all of the following conditions: 
(i) $-1 <x_i <1,$ for all $1\leq  i \leq  n.$
(ii) $ x_1 + x_2 + ... + x_n = 0.$
(iii) $\sqrt{1 - x_1^2} +\sqrt{1 - x^2_2} + ... +\sqrt{1 - x^2_n} = 1.$
1. **Case \( n = 1 \):**
   - If \( n = 1 \), we need to find \( x_1 \) such that \( -1 < x_1 < 1 \), \( x_1 = 0 \), and \( \sqrt{1 - x_1^2} = 1 \).
   - Choosing \( x_1 = 0 \) satisfies all conditions:
     \[
     -1 < 0 < 1, \quad 0 = 0, \quad \sqrt{1 - 0^2} = 1.
     \]
   - Therefore, \( n = 1 \) is a solution.

2. **Case \( n = 2k \) (even \( n \)):**
   - Let \( n = 2k \). We need to find \( x_1, x_2, \ldots, x_{2k} \) such that:
     \[
     -1 < x_i < 1 \quad \text{for all } i,
     \]
     \[
     x_1 + x_2 + \cdots + x_{2k} = 0,
     \]
     \[
     \sqrt{1 - x_1^2} + \sqrt{1 - x_2^2} + \cdots + \sqrt{1 - x_{2k}^2} = 1.
     \]
   - Choose \( x_1 = x_2 = \cdots = x_k = \sqrt{1 - \frac{1}{k^2}} \) and \( x_{k+1} = x_{k+2} = \cdots = x_{2k} = -\sqrt{1 - \frac{1}{k^2}} \).
   - Verify the conditions:
     \[
     -1 < \sqrt{1 - \frac{1}{k^2}} < 1 \quad \text{and} \quad -1 < -\sqrt{1 - \frac{1}{k^2}} < 1,
     \]
     \[
     k \cdot \sqrt{1 - \frac{1}{k^2}} + k \cdot \left(-\sqrt{1 - \frac{1}{k^2}}\right) = 0,
     \]
     \[
     k \cdot \sqrt{1 - \left(\sqrt{1 - \frac{1}{k^2}}\right)^2} + k \cdot \sqrt{1 - \left(-\sqrt{1 - \frac{1}{k^2}}\right)^2} = k \cdot \sqrt{\frac{1}{k^2}} + k \cdot \sqrt{\frac{1}{k^2}} = k \cdot \frac{1}{k} + k \cdot \frac{1}{k} = 1.
     \]
   - Therefore, \( n = 2k \) (even \( n \)) is a solution.

3. **Case \( n = 2k + 1 \) (odd \( n \)):**
   - Let \( n = 2k + 1 \). We need to show that no such \( x_1, x_2, \ldots, x_{2k+1} \) exist.
   - Assume \( x_i \neq 0 \) for any \( i \), otherwise:
     \[
     \sqrt{1 - x_1^2} + \sqrt{1 - x_2^2} + \cdots + \sqrt{1 - x_{2k+1}^2} > 1.
     \]
   - This forces at least \( k+1 \) of \( \{x_1, x_2, \ldots, x_{2k+1}\} \) to have the same sign. Without loss of generality, assume \( x_1, x_2, \ldots, x_{k+1} \) are all positive.
   - For any \( x \in (0, 1) \), the inequality \( x > 1 - \sqrt{1 - x^2} \) holds. Therefore:
     \[
     x_1 + x_2 + \cdots + x_{k+1} > (k+1) - \sum_{i=1}^{k+1} \sqrt{1 - x_i^2} > k.
     \]
   - Hence:
     \[
     x_1 + x_2 + \cdots + x_{2k+1} > k + (x_{k+2} + x_{k+3} + \cdots + x_{2k+1}) > 0,
     \]
     which contradicts the condition \( x_1 + x_2 + \cdots + x_{2k+1} = 0 \).

Therefore, the only possible values of \( n \) are \( n = 1 \) and all even \( n \).

The final answer is \( \boxed{ n = 1 } \) and all even \( n \).