Find all numbers $n \in \mathbb{N}^*$ for which there exists a finite set of natural numbers $A=(a_1, a_2,...a_n)$ so that for any $k$ $(1\leq k \leq n)$ the number $a_k$ is the number of all multiples of $k$ in set $A$.
1. **Base Cases:**
   - For \( n = 1 \):
     - The set \( A = \{1\} \) satisfies the condition because \( a_1 = 1 \) and there is exactly one multiple of 1 in \( A \).
   - For \( n = 2 \):
     - The set \( A = \{2, 1\} \) satisfies the condition because \( a_1 = 2 \) (both 1 and 2 are multiples of 1) and \( a_2 = 1 \) (only 2 is a multiple of 2).

2. **General Case for \( n > 2 \):**
   - Assume \( n > 2 \).
   - Note that \( 0 \not\in A \) and \( a_1 = n \) because every number in \( A \) is a multiple of 1.
   - Suppose \( a_i \neq a_j \) for every pair \( (i, j) \) such that \( i \neq j \). This implies \( a_1, a_2, \ldots, a_n \) is a permutation of \( 1, 2, \ldots, n \).

3. **Contradiction for \( n > 2 \):**
   - For \( n > 2 \), there is only one number in the set \( \{1, 2, \ldots, n\} \) that is divisible by \( n-1 \). This implies \( a_{n-1} = 1 \).
   - However, if \( a_{n-1} = 1 \), then \( a_{n-1} = a_n = 1 \), which contradicts the assumption that \( a_i \neq a_j \) for \( i \neq j \).

4. **Considering Equal Numbers:**
   - If numbers can be equal, then \( a_{n-1} = 1 \) because \( 0 < a_i < n+1 \).
   - There must exist an \( i \) such that \( a_i = n-1 \). This means \( i \) divides \( n-1 \) numbers in \( A \), but since two of them are equal to 1, \( a_i \leq n-2 \), which is a contradiction.

5. **Conclusion:**
   - We have shown that there is no solution for \( n > 2 \).

The final answer is \( \boxed{ n = 1 } \) and \( n = 2 \).