For which $ n\in \mathbb{N}$ do there exist rational numbers $ a,b$ which are not integers such that both $ a \plus{} b$ and $ a^n \plus{} b^n$ are integers?
To determine for which \( n \in \mathbb{N} \) there exist rational numbers \( a \) and \( b \) which are not integers such that both \( a + b \) and \( a^n + b^n \) are integers, we proceed as follows:

1. **Express \( a \) and \( b \) as rational numbers:**
   Let \( a = \frac{x}{p} \) and \( b = \frac{y}{q} \), where \( \gcd(x, p) = 1 \) and \( \gcd(y, q) = 1 \), and \( p, q > 0 \).

2. **Condition for \( a + b \) to be an integer:**
   \[
   a + b = \frac{x}{p} + \frac{y}{q} = \frac{xq + yp}{pq}
   \]
   For \( a + b \) to be an integer, \( pq \) must divide \( xq + yp \). This implies:
   \[
   pq \mid xq + yp
   \]
   Since \( \gcd(x, p) = 1 \) and \( \gcd(y, q) = 1 \), it follows that \( p \mid q \) and \( q \mid p \). Therefore, \( p = q \).

3. **Simplify the problem:**
   Now, we only need to consider \( a = \frac{x}{p} \) and \( b = \frac{y}{p} \) with \( \gcd(x, y) = 1 \). The condition becomes:
   \[
   p \mid x + y
   \]
   and
   \[
   p^n \mid x^n + y^n
   \]

4. **Case 1: \( n \) is even:**
   Let \( n = 2k \). We need \( p^n \mid x^n + y^n \). Since \( \gcd(x + y, x^n + y^n) \in \{1, 2\} \), it follows that \( p = 2 \). However, if \( p = 2 \), then \( 2^n \mid x^n + y^n \) implies \( 2 \mid x \) and \( 2 \mid y \), which contradicts \( \gcd(x, y) = 1 \). Therefore, there are no solutions for even \( n \).

5. **Case 2: \( n \) is odd:**
   Let \( n = 2k + 1 \). We need \( p^n \mid x^n + y^n \). Consider a prime \( p \) dividing \( x + y \). Using the p-adic valuation \( v_p \), we have:
   \[
   v_p(x^n + y^n) = v_p(x + y) + v_p(n)
   \]
   If we choose \( v_p(x + y) \geq v_p(n - 1) \), then the condition is satisfied. Therefore, there exist rational numbers \( a \) and \( b \) such that \( a + b \) and \( a^n + b^n \) are integers for odd \( n \).

Conclusion:
The condition is satisfied if and only if \( n \) is odd.

The final answer is \( \boxed{ n \equiv 1 \pmod{2} } \).