find all pairs of relatively prime natural numbers $ (m,n) $ in such a way that there exists non constant polynomial f satisfying \[ gcd(a+b+1, mf(a)+nf(b)  > 1 \]
for every natural numbers $ a $ and $ b $
1. **Initial Assumption and Example:**
   - We are given that for \( m = n = 1 \), the polynomial \( f(x) = 2x + 1 \) satisfies the condition.
   - We need to find all pairs of relatively prime natural numbers \( (m, n) \) such that there exists a non-constant polynomial \( f \) satisfying:
     \[
     \gcd(a + b + 1, mf(a) + nf(b)) > 1
     \]
     for every natural number \( a \) and \( b \).

2. **Assume \( mn \neq 1 \):**
   - Suppose such a polynomial \( f \) exists. Pick a natural number \( t \) for which \( f(t) \neq 0 \).
   - Set \( a = u - t - 1 \) and \( b = u^2 - u + t \), where \( u \) is an arbitrary integer larger than \( t + 1 \).

3. **Substitution and Modulo Calculation:**
   - Calculate \( mf(a) + nf(b) \):
     \[
     mf(a) + nf(b) = mf(u - t - 1) + nf(u^2 - u + t)
     \]
   - Consider the expression modulo \( u \):
     \[
     mf(u - t - 1) + nf(u^2 - u + t) \equiv mf(-t - 1) + nf(t) \pmod{u}
     \]
   - Let \( k = mf(-t - 1) + nf(t) \). We assume \( k \neq 0 \) since both \( mf(-t - 1) \) and \( nf(t) \) cannot be zero simultaneously.

4. **Choosing \( u \) and Ensuring Coprimality:**
   - Since \( a + b + 1 = u^2 \), we have:
     \[
     \gcd(u^2, k) > 1
     \]
   - We can choose \( u \) such that \( k \) is coprime to \( u \). This implies that \( k \) must be a multiple of some prime factor of \( u \).

5. **Conclusion:**
   - The only way to ensure that \( \gcd(a + b + 1, mf(a) + nf(b)) > 1 \) for all \( a \) and \( b \) is if \( m = n = 1 \).
   - Therefore, the only solution is \( (m, n) = (1, 1) \).

The final answer is \( \boxed{ (m, n) = (1, 1) } \).