Call a two-element subset of $\mathbb{N}$ [i]cute[/i] if it contains exactly one prime number and one composite number. Determine all polynomials $f \in \mathbb{Z}[x]$ such that for every [i]cute[/i] subset $ \{ p,q \}$, the subset $ \{ f(p) + q, f(q) + p \} $ is [i]cute[/i] as well.

[i]Proposed by Valentio Iverson (Indonesia)[/i]
1. **Assume \( f \) is non-constant:**
   - If \( f \) is constant, the only polynomial in \(\mathbb{Z}[x]\) that satisfies the condition is \( f(x) = 0 \). This is because for any cute subset \(\{p, q\}\), \(\{f(p) + q, f(q) + p\} = \{q, p\}\) is also cute. Hence, \( f(x) = 0 \) works.

2. **Consider the constant coefficient of \( f \):**
   - Let \( f(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 \).
   - If \( 2 \) divides the constant coefficient \( a_0 \), then consider \( p = 2 \) and \( q = 2k \) for \( k > 1 \). 
     - We need \(\{f(2) + 2k, f(2k) + 2\}\) to be cute.
     - Since \( f(2) \) and \( f(2k) \) are both integers, and \( 2 \) divides \( a_0 \), both \( f(2) \) and \( f(2k) \) are even.
     - Therefore, \( f(2) + 2k \) and \( f(2k) + 2 \) are both even, making them composite for sufficiently large \( k \). This contradicts the requirement that the subset is cute.

3. **Consider the case where \( 2 \) does not divide the constant coefficient \( a_0 \):**
   - Let \( q = 2f(p) \).
   - We need \(\{f(p) + 2f(p), f(2f(p)) + p\} = \{3f(p), f(2f(p)) + p\}\) to be cute.
   - Since \( f \) is non-constant, \( 3f(p) \) will eventually be composite for sufficiently large primes \( p \).
   - This implies \( f(2f(p)) + p \) must be prime.
   - However, since \( f(2f(p)) \) and \( p \) are both odd, \( f(2f(p)) + p \) is even.
   - The only even prime is \( 2 \), so \( f(2f(p)) + p = 2 \).
   - This implies \( f(2f(p)) = 2 - p \), which is not possible for infinitely many primes \( p \).

4. **Conclusion:**
   - The only polynomial \( f \in \mathbb{Z}[x] \) that satisfies the given condition is \( f(x) = 0 \).

The final answer is \( \boxed{ f(x) = 0 } \).