Find all functions $f : \mathbb N \to \mathbb N$ such that:

i) $f^{2000}(m)=f(m)$ for all $m \in \mathbb N$,

ii) $f(mn)=\dfrac{f(m)f(n)}{f(\gcd(m,n))}$, for all $m,n\in \mathbb N$, and

iii) $f(m)=1$ if and only if $m=1$.
1. **Multiplicative Property:**
   Since \( f \) is multiplicative for coprime numbers, we can write:
   \[
   f(mn) = f(m) f(n) \quad \text{if} \quad \gcd(m, n) = 1
   \]

2. **Behavior on Powers of Primes:**
   Let's consider \( f(p^k) \) for a prime \( p \) and \( k \in \mathbb{N} \). Using the multiplicative property and condition (ii), we get:
   \[
   f(p^k) = f(p \cdot p^{k-1}) = \frac{f(p) f(p^{k-1})}{f(\gcd(p, p^{k-1}))} = \frac{f(p) f(p^{k-1})}{f(p)} = f(p^{k-1})
   \]
   By induction, we can show that:
   \[
   f(p^k) = f(p)
   \]
   for all \( k \in \mathbb{N} \).

3. **General Form for Composite Numbers:**
   Consider a composite number \( m \) with prime factorization \( m = \prod_{i=1}^k p_i^{a_i} \). Using the multiplicative property and the result for powers of primes, we get:
   \[
   f(m) = f\left(\prod_{i=1}^k p_i^{a_i}\right) = \prod_{i=1}^k f(p_i^{a_i}) = \prod_{i=1}^k f(p_i)
   \]

4. **Periodic Behavior:**
   From condition (i), \( f \) has a periodic behavior with period dividing 2000. This means that for any prime \( p \), \( f(p) \) must be such that \( f^{2000}(p) = f(p) \).

5. **Uniqueness and Form of \( f \):**
   Given the periodicity and the multiplicative property, we can conclude that \( f \) must map each prime \( p \) to a fixed value \( f(p) \) that is a power of a prime. Specifically, we can write:
   \[
   f(p_i) = p_{G(i)}^{I(p_i)}
   \]
   where \( G \) is a permutation of the set of primes with no cycles smaller than 1999, and \( I \) is a function mapping primes to natural numbers.

6. **Conclusion:**
   The function \( f \) can be defined as:
   \[
   f\left(\prod_{i} p_{i}^{a_{i}}\right) = \prod_{i} p_{G(i)}^{H(a_{i})I(p_{i})}
   \]
   where \( H \) maps 0 to 0 and natural numbers to 1, \( G \) is a permutation of primes with no cycles smaller than 1999, and \( I \) maps primes to natural numbers.

The final answer is \( \boxed{ f\left(\prod_{i} p_{i}^{a_{i}}\right) = \prod_{i} p_{G(i)}^{H(a_{i})I(p_{i})} } \).