Find all positive integers $n$ such that there exists an infinite set $A$ of positive integers with the following property: For all pairwise distinct numbers $a_1, a_2, \ldots , a_n \in A$, the numbers
$$a_1 + a_2 + \ldots + a_n \text{ and } a_1\cdot a_2\cdot \ldots\cdot a_n$$
are coprime.
1. **Case for \( n = 1 \):**
   - If \( n = 1 \), we need to find a set \( A \) such that for any \( a_1 \in A \), the number \( a_1 \) is coprime with itself. This is impossible because any number is not coprime with itself. Hence, there is no such set \( A \) for \( n = 1 \).

2. **Case for \( n \geq 2 \):**
   - We need to construct an infinite set \( A \) of positive integers such that for any \( n \) pairwise distinct numbers \( a_1, a_2, \ldots, a_n \in A \), the sum \( a_1 + a_2 + \ldots + a_n \) and the product \( a_1 \cdot a_2 \cdot \ldots \cdot a_n \) are coprime.

3. **Choosing primes:**
   - Let \( p_1 \) be a prime number greater than \( n \).
   - For \( i \geq 2 \), choose \( p_i \) such that:
     \[
     p_i \equiv 1 \pmod{p_1 \cdot p_2 \cdots p_{i-1}} \quad \text{and} \quad p_i > p_1 + p_2 + \cdots + p_{i-1} + n
     \]
     By Dirichlet's theorem on arithmetic progressions, such primes \( p_i \) exist.

4. **Constructing the set \( A \):**
   - Let \( A = \{ p_1, p_2, p_3, \ldots \} \).

5. **Verifying the coprimality condition:**
   - Consider any \( n \) pairwise distinct numbers \( q_1, q_2, \ldots, q_n \in A \) with \( q_1 < q_2 < \cdots < q_n \).
   - Assume there exists a prime \( r \) such that \( r \) divides both \( q_1 + q_2 + \cdots + q_n \) and \( q_1 \cdot q_2 \cdot \cdots \cdot q_n \).
   - Since \( r \) divides \( q_1 \cdot q_2 \cdot \cdots \cdot q_n \), \( r \) must be one of the \( q_i \), say \( r = q_t \) for some \( t \in \{1, 2, \ldots, n\} \).

6. **Deriving a contradiction:**
   - Since \( q_t \) divides \( q_1 + q_2 + \cdots + q_n \), we have:
     \[
     q_t \mid q_1 + q_2 + \cdots + q_{t-1} + q_{t+1} + \cdots + q_n
     \]
   - Given the construction of \( p_i \), we have \( q_{t+1} \equiv q_{t+2} \equiv \cdots \equiv q_n \equiv 1 \pmod{q_t} \).
   - Therefore, \( q_t \mid q_1 + q_2 + \cdots + q_{t-1} + (n - t) \).

7. **Bounding \( q_t \):**
   - Let \( Q_t = \{ x \in A : x < q_t \} \). Clearly, \( q_1, q_2, \ldots, q_{t-1} \in Q_t \).
   - By construction, \( q_t > (\text{sum of the elements of } Q_t) + n \geq q_1 + q_2 + \cdots + q_{t-1} + n \).
   - This implies \( q_t > q_1 + q_2 + \cdots + q_{t-1} + (n - t) \), which contradicts the earlier condition that \( q_t \mid q_1 + q_2 + \cdots + q_{t-1} + (n - t) \).

8. **Conclusion:**
   - The contradiction implies that no such prime \( r \) exists, and thus \( q_1 + q_2 + \cdots + q_n \) and \( q_1 \cdot q_2 \cdot \cdots \cdot q_n \) are coprime.

The final answer is all positive integers except \( \boxed{ n = 1 } \).