Problem 1299

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 $\boxed{ n = 1 }$.