Find all functions $ f: \mathbb{N} \to \mathbb{Z} $ satisfying $$ n \mid f\left(m\right) \Longleftrightarrow m \mid \sum\limits_{d \mid n}{f\left(d\right)} $$ holds for all positive integers $ m,n $
1. **Claim 1:** \( n \mid f(m) \Longleftrightarrow m \mid f(n) \)

   **Proof:**
   Let \( d \) be a proper divisor of \( n \). We start with the given condition:
   \[
   m \mid \sum_{d \mid n} f(d) \Longleftrightarrow n \mid f(m)
   \]
   This implies:
   \[
   d \mid f(m) \Longleftrightarrow m \mid \sum_{d' \mid d} f(d')
   \]
   By iterating this process infinitely many times, we get:
   \[
   n \mid f(m) \Longleftrightarrow m \mid f(n)
   \]
   \(\blacksquare\)

2. From Claim 1, we know that \( 1 \mid f(n) \) for any \( n \). This implies that \( f(n) \) must be divisible by 1, which is always true. However, it also implies that \( n \mid f(1) \) for any \( n \). Therefore, \( f(1) \) must be 0 because the only integer divisible by all positive integers is 0. Hence, \( f(1) = 0 \).

3. **Claim 2:** \( f(nk) \mid f(n) \) or \( f(nk) = 0 \)

   **Proof:**
   Consider the given condition:
   \[
   m \mid f(nk) \Longleftrightarrow nk \mid f(m)
   \]
   This implies:
   \[
   n \mid f(m) \Longleftrightarrow m \mid f(n)
   \]
   By Claim 1, we know that \( n \mid f(m) \Longleftrightarrow m \mid f(n) \). Therefore, we conclude that:
   \[
   f(nk) \mid f(n) \text{ or } f(nk) = 0
   \]
   \(\blacksquare\)

4. Suppose there exists a prime \( p \) such that \( f(p) \neq 0 \). From Claim 2, there exists an integer \( \alpha \) such that \( p^{\alpha} \) does not divide any \( f(n) \). However, from Claim 1, we have:
   \[
   p^{\alpha} \mid f(f(p^{\alpha}))
   \]
   This leads to a contradiction because if \( p^{\alpha} \) does not divide any \( f(n) \), it cannot divide \( f(f(p^{\alpha})) \). Therefore, our assumption that \( f(p) \neq 0 \) must be false. Hence, \( f(p) = 0 \) for any prime \( p \).

5. Since \( f(p) = 0 \) for any prime \( p \), and by Claim 2, \( f(n) = 0 \) for any \( n \) that is a product of primes, we conclude that \( f(n) = 0 \) for all \( n \in \mathbb{N} \).

The final answer is \( \boxed{ f(n) = 0 } \) for all \( n \in \mathbb{N} \).