$\mathbb{N}$ is the set of positive integers. Determine all functions $f:\mathbb{N}\to\mathbb{N}$ such that for every pair $(m,n)\in\mathbb{N}^2$ we have that:
\[f(m)+f(n) \ | \ m+n .\]
1. Let \( P(m, n) \) denote the assertion \( f(m) + f(n) \mid m + n \).

2. Consider \( P(1, 1) \):
   \[
   f(1) + f(1) \mid 1 + 1 \implies 2f(1) \mid 2.
   \]
   Since \( f(1) \) is a positive integer, the only solution is \( f(1) = 1 \).

3. Next, consider \( P(2, 1) \):
   \[
   f(2) + f(1) \mid 2 + 1 \implies f(2) + 1 \mid 3.
   \]
   The possible values for \( f(2) \) are 2, since \( f(2) + 1 \) must be a divisor of 3. Thus, \( f(2) = 2 \).

4. Now, pick an odd prime \( p \). Consider \( P(p-1, 1) \):
   \[
   f(p-1) + f(1) \mid p-1 + 1 \implies f(p-1) + 1 \mid p.
   \]
   Since \( p \) is a prime number, the possible values for \( f(p-1) + 1 \) are 1 and \( p \). However, \( f(p-1) + 1 = 1 \) is not possible because \( f(p-1) \) is a positive integer. Therefore, \( f(p-1) + 1 = p \), which implies \( f(p-1) = p-1 \).

5. We have shown that for any odd prime \( p \), \( f(p-1) = p-1 \). This suggests that there exists a large \( m \) such that \( f(m) = m \).

6. Consider \( P(m, n) \) for arbitrary \( m \) and \( n \):
   \[
   f(m) + f(n) \mid m + n.
   \]
   If \( f(m) = m \) for sufficiently large \( m \), then:
   \[
   m + f(n) \mid m + n \implies m + f(n) \mid f(n) - n.
   \]
   Fix \( n \) and take \( m \) sufficiently large. Since \( m + f(n) \) must divide \( f(n) - n \), and \( m \) can be arbitrarily large, the only way this can hold is if \( f(n) = n \).

7. Therefore, the only function that satisfies the given condition for all \( m, n \in \mathbb{N} \) is \( f(n) = n \).

The final answer is \( \boxed{f(n) = n \text{ for all } n \in \mathbb{N}} \).