We are tasked with finding all surjective functions $f: \mathbb{N} \to \mathbb{N}$ that satisfy the condition: for every $m, n \in \mathbb{N}$ and every prime $p$, the number $f(m+n)$ is divisible by $p$ if and only if $f(m) + f(n)$ is divisible by $p$.

To solve this, we consider the given condition:

$$p \mid f(m+n) \iff p \mid (f(m) + f(n)).$$

Let's explore the implications of this condition:

1. Injectivity:
Assume for contradiction that $f(a) = f(b)$ for $a \neq b$. Then, consider $m = a$ and $n = 0$:
$$f(a+0) \equiv f(b+0) \equiv f(a) + f(0) = f(b) + f(0).$$
Since $f$ is assumed to be surjective, it must be injective as well because if $f(a) = f(b)$, any number $c$ in the codomain cannot have two different pre-images, which would violate surjectivity.

2. Additivity and Linear Form:
For simplicity, consider $m = n = 1$:
$$f(2) \equiv 2f(1).$$
Now generalize this idea. Suppose by induction that $f(k) = k \cdot f(1)$ holds for some $k$. Then, for $m = k$ and $n = 1$:
$$f(k+1) \equiv f(k) + f(1) = kf(1) + f(1) = (k+1)f(1),$$
thereby maintaining the linearity $f(n) = nf(1)$.

3. Scaling:
Consider $a = 1$, then $f(1+1)=2f(1)$ should hold. Scaling continues to suggest that $f(n) = nf(1)$ and let's assume $f(1) = 1$ for surjectivity $f(n) = n$.

4. Testing the Condition:
Given $f(n) = n$, check the condition in both directions:

- If $n+m = kp$, then $f(n+m) = n + m \equiv 0 \pmod{p}$.
- If $f(n) + f(m) = n + m = kp$, similarly $f(n+m) = kp \equiv 0 \pmod{p}$.

The only function which satisfies all constraints and maintain surjectivity is $f(n) = n$.

Thus, the function that satisfies the given condition is:
$$\boxed{f(n) = n}.$$