For which positive integers $n$, one can find a non-integer rational number $x$ such that $$x^n+(x+1)^n$$ is an integer?
1. **Case \( n = 1 \):**
   - If \( n = 1 \), we can choose \( x = \frac{1}{2} \).
   - Then, \( x^1 + (x+1)^1 = \frac{1}{2} + \left(\frac{1}{2} + 1\right) = \frac{1}{2} + \frac{3}{2} = 2 \), which is an integer.
   - Therefore, \( x = \frac{1}{2} \) works for \( n = 1 \).

2. **Case \( n > 1 \):**
   - Let \( x = \frac{a}{b} \) where \( a, b \in \mathbb{Z} \), \( b > 1 \), and \(\gcd(a, b) = 1 \).
   - We need \( x^n + (x+1)^n \) to be an integer, i.e., 
     \[
     \left(\frac{a}{b}\right)^n + \left(\frac{a+b}{b}\right)^n \in \mathbb{Z}
     \]
   - This implies:
     \[
     \frac{a^n + (a+b)^n}{b^n} \in \mathbb{Z} \implies b^n \mid a^n + (a+b)^n
     \]
   - Since \( b^n \mid a^n + (a+b)^n \), it follows that \( b \mid a^n + (a+b)^n \).

3. **Divisibility Analysis:**
   - Since \( b \mid a^n + (a+b)^n \), we have:
     \[
     b \mid a^n + (a^n + n a^{n-1} b + \cdots + b^n) = 2a^n + n a^{n-1} b + \cdots + b^n
     \]
   - For \( b \) to divide this expression, \( b \) must divide each term. In particular, \( b \mid 2a^n \).
   - Since \( \gcd(a, b) = 1 \), \( b \) must divide 2. Therefore, \( b = 2 \).

4. **Case \( b = 2 \):**
   - If \( b = 2 \), then \( x = \frac{a}{2} \) where \( \gcd(a, 2) = 1 \) (i.e., \( a \) is odd).
   - We need to check when \( \left(\frac{a}{2}\right)^n + \left(\frac{a+2}{2}\right)^n \) is an integer.

5. **Even \( n \):**
   - If \( n \) is even, then \( 2^n \mid a^n + (a+2)^n \).
   - Consider \( a^n \mod 4 \) and \( (a+2)^n \mod 4 \):
     - Since \( a \) is odd, \( a \equiv 1 \) or \( 3 \mod 4 \).
     - If \( a \equiv 1 \mod 4 \), then \( a^n \equiv 1 \mod 4 \) and \( (a+2)^n \equiv 3^n \equiv 1 \mod 4 \) (since \( n \) is even).
     - If \( a \equiv 3 \mod 4 \), then \( a^n \equiv 3^n \equiv 1 \mod 4 \) and \( (a+2)^n \equiv 1^n \equiv 1 \mod 4 \).
   - In both cases, \( a^n + (a+2)^n \equiv 2 \mod 4 \), which is not divisible by \( 4 \). Hence, \( n \) cannot be even.

6. **Odd \( n \):**
   - If \( n \) is odd, consider \( a = 2^{n-1} - 1 \) and \( b = 2 \):
     \[
     x = \frac{2^{n-1} - 1}{2}
     \]
   - Then,
     \[
     x^n + (x+1)^n = \left(\frac{2^{n-1} - 1}{2}\right)^n + \left(\frac{2^{n-1} - 1 + 2}{2}\right)^n = \left(\frac{2^{n-1} - 1}{2}\right)^n + \left(\frac{2^{n-1} + 1}{2}\right)^n
     \]
   - Simplifying,
     \[
     \left(\frac{2^{n-1} - 1}{2}\right)^n + \left(\frac{2^{n-1} + 1}{2}\right)^n = \frac{(2^{n-1} - 1)^n + (2^{n-1} + 1)^n}{2^n}
     \]
   - Since \( n \) is odd, the binomial expansion of \( (2^{n-1} - 1)^n \) and \( (2^{n-1} + 1)^n \) will have terms that cancel out, leaving a multiple of \( 2^n \) in the numerator. Thus, the expression is an integer.

Conclusion:
- For \( n = 1 \), \( x = \frac{1}{2} \) works.
- For \( n > 1 \), \( n \) must be odd, and \( x = \frac{2^{n-1} - 1}{2} \) works.

The final answer is \( \boxed{ n } \) must be odd.