Determine all positive integers $n$ for which the equation 
\[ x^n + (2+x)^n + (2-x)^n = 0 \] 
has an integer as a solution.
1. **Case 1: \( n \) is even**

   If \( n \) is even, then each term \( x^n \), \((2+x)^n\), and \((2-x)^n\) is non-negative because raising any real number to an even power results in a non-negative number. Therefore, we have:
   \[
   x^n + (2+x)^n + (2-x)^n \geq 0
   \]
   Since the sum of non-negative numbers is non-negative, the equation \( x^n + (2+x)^n + (2-x)^n = 0 \) can only hold if each term is zero. However, this would require:
   \[
   x^n = 0, \quad (2+x)^n = 0, \quad (2-x)^n = 0
   \]
   The only way for \( x^n = 0 \) is if \( x = 0 \). But then \( (2+0)^n = 2^n \neq 0 \) and \( (2-0)^n = 2^n \neq 0 \). Thus, there are no solutions for even \( n \).

2. **Case 2: \( n \) is odd**

   If \( n \) is odd, the equation \( x^n + (2+x)^n + (2-x)^n = 0 \) can be rewritten as:
   \[
   x^n + (2+x)^n = -(2-x)^n
   \]
   Since \( n \) is odd, \((2-x)^n\) will have the same sign as \( 2-x \). We need to check specific values of \( n \).

   - **For \( n = 1 \)**:
     \[
     x + (2+x) + (2-x) = 0
     \]
     Simplifying, we get:
     \[
     x + 2 + x + 2 - x = 0 \implies 2x + 4 = 0 \implies x = -2
     \]
     This is a valid integer solution.

   - **For \( n \geq 3 \)**:
     Consider \( n = 3 \):
     \[
     x^3 + (2+x)^3 + (2-x)^3 = 0
     \]
     Expanding the terms:
     \[
     x^3 + (8 + 12x + 6x^2 + x^3) + (8 - 12x + 6x^2 - x^3) = 0
     \]
     Simplifying, we get:
     \[
     x^3 + x^3 + 8 + 12x + 6x^2 + 8 - 12x + 6x^2 - x^3 = 0 \implies 2x^3 + 12x^2 + 16 = 0
     \]
     This simplifies to:
     \[
     2x^3 + 12x^2 + 16 = 0 \implies x^3 + 6x^2 + 8 = 0
     \]
     Solving this cubic equation for integer solutions, we find that there are no integer solutions for \( x \).

   By Fermat's Last Theorem, for \( n \geq 3 \), there are no non-trivial integer solutions to the equation \( x^n + y^n = z^n \) for \( n \geq 3 \). Therefore, there are no integer solutions for \( n \geq 3 \).

Conclusion:
The only valid \( n \) for which the equation has integer solutions in \( x \) is \( \boxed{n=1} \).