Consider the functional equation where $f : \mathbb{N} \rightarrow \mathbb{N}$ satisfies, for all positive integers $m$ and $n$,

$$f(m) + n - m \equiv 0 \pmod{f(n)}.$$

This condition implies:

$$f(m) + n \equiv m \pmod{f(n)}.$$

We aim to find all such functions $f$.

### Case 1: Linear Functions of the Form $f(n) = n + c$

Assume $f(n) = n + c$ for some constant $c \in \mathbb{N}$. Substituting into the original equation, we have:

$$f(m) + n - m = (m + c) + n - m = n + c.$$

This implies:

$$n + c \equiv m \pmod{n + c}.$$

The left-hand side is divisible by $n + c$ since:

$$n + c - m \equiv 0 \pmod{n + c},$$

which is trivially true for any integer $m$. Thus, $f(n) = n + c$ is a valid solution.

### Case 2: Constant Function $f(n) \equiv 1$

Assume $f(n) \equiv 1$. Substituting into the original equation, we have:

$$f(m) + n - m \equiv 0 \pmod{1},$$

which simplifies to:

$$(m + n - m) \equiv 0 \pmod{1},$$

which is always true. Thus, $f(n) = 1$ is a valid solution.

### Case 3: Piecewise Function $f(\text{even}) = 1, f(\text{odd}) = 2$

Consider $f(n) = 1$ for even $n$ and $f(n) = 2$ for odd $n$.

1. **$n$ is even**: $f(n) = 1$. The equation becomes:

$$f(m) + n - m \equiv 0 \pmod{1},$$

which holds true since any number is divisible by 1.

2. **$n$ is odd**: $f(n) = 2$. The equation becomes:

$$f(m) + n - m \equiv 0 \pmod{2}.$$

For any $m$ and odd $n$, $n - m \equiv 0 \pmod{2}$ as both $n$ and $m$ would have the same parity.

Thus, $f(\text{even}) = 1, f(\text{odd}) = 2$ is a valid solution.

### Case 4: Piecewise Function $f(\text{odd}) = 1, f(\text{even}) = 2$

Consider the similar argument for the reverse configuration:

1. **$n$ is odd**: $f(n) = 1$. The equation becomes:

$$f(m) + n - m \equiv 0 \pmod{1},$$

which holds true.

2. **$n$ is even**: $f(n) = 2$. The equation becomes:

$$f(m) + n - m \equiv 0 \pmod{2}.$$

For any $m$ and even $n$, both $n$ and $m$ would have the same parity, thus maintaining divisibility.

Hence, $f(\text{odd}) = 1, f(\text{even}) = 2$ is also a valid solution.

### Conclusion

The functions that satisfy the given conditions are:

$$\boxed{f(n) = n + c}, \boxed{f(n) \equiv 1}, \boxed{f(\text{even}) = 1, f(\text{odd}) = 2}, \boxed{f(\text{odd}) = 1, f(\text{even}) = 2}.$$