Find all positive integers $n$ for which exist three nonzero integers $x, y, z$ such that $x+y+z=0$ and:
\[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{n}\]
We are given the constraints \( x + y + z = 0 \) and 
\[
\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{n}.
\]

1. **Substitute \( z = -x - y \) into the second equation:**
   \[
   \frac{1}{x} + \frac{1}{y} + \frac{1}{-x - y} = \frac{1}{n}.
   \]

2. **Combine the fractions on the left-hand side:**
   \[
   \frac{xy(-x - y) + x(-x - y) + y(-x - y)}{xy(-x - y)} = \frac{1}{n}.
   \]

3. **Simplify the numerator:**
   \[
   xy(-x - y) + x(-x - y) + y(-x - y) = -x^2y - xy^2 - x^2 - xy - yx - y^2 = -x^2 - xy - y^2.
   \]

4. **Simplify the entire fraction:**
   \[
   \frac{-x^2 - xy - y^2}{xy(-x - y)} = \frac{1}{n}.
   \]

5. **Simplify further:**
   \[
   \frac{x^2 + xy + y^2}{xy(x + y)} = \frac{1}{n}.
   \]

6. **Rearrange to solve for \( n \):**
   \[
   n = \frac{xy(x + y)}{x^2 + xy + y^2}.
   \]

7. **Consider the case \( x = y \):**
   \[
   n = \frac{x^2(x + x)}{x^2 + x^2 + x^2} = \frac{2x^3}{3x^2} = \frac{2x}{3}.
   \]
   For \( n \) to be an integer, \( x \) must be a multiple of 3. Let \( x = 3k \), then:
   \[
   n = \frac{2(3k)}{3} = 2k.
   \]
   Thus, \( n \) can be any even integer.

8. **Show that odd \( n \) is not achievable:**
   - If \( x \) is odd and \( y \) is even, or vice versa, the numerator \( xy(x + y) \) is even, and the denominator \( x^2 + xy + y^2 \) is odd, making \( n \) non-integer.
   - If both \( x \) and \( y \) are odd, the numerator is even (since \( x + y \) is even) and the denominator is odd, making \( n \) non-integer.
   - If both \( x \) and \( y \) are even, let \( x = 2m \) and \( y = 2n \):
     \[
     n = \frac{(2m)(2n)(2m + 2n)}{(2m)^2 + (2m)(2n) + (2n)^2} = \frac{4mn(2m + 2n)}{4(m^2 + mn + n^2)} = \frac{2mn(m + n)}{m^2 + mn + n^2}.
     \]
     This reduces to a multiple of 2, thus \( n \) is even.

Therefore, we have shown that only even \( n \) is possible.

The final answer is \( \boxed{ n = 2k } \) for all \( k \in \mathbb{Z} \setminus \{0\} \).