For which positive integers $n$ does there exist a positive integer $m$ such that among the numbers $m + n, 2m + (n - 1), \dots, nm + 1$, there are no two that share a common factor greater than $1$?
To determine for which positive integers \( n \) there exists a positive integer \( m \) such that among the numbers \( m + n, 2m + (n - 1), \dots, nm + 1 \), there are no two that share a common factor greater than 1, we proceed as follows:

1. **Define the sequence:**
   Let \( a_i = i \cdot m + n + 1 - i \) for \( i \leq n \). This sequence represents the numbers \( m + n, 2m + (n - 1), \dots, nm + 1 \).

2. **Calculate the greatest common divisor (GCD):**
   Consider two terms \( a_i \) and \( a_j \) where \( i < j \). We need to find \(\gcd(a_i, a_j)\):
   \[
   a_i = i \cdot m + n + 1 - i
   \]
   \[
   a_j = j \cdot m + n + 1 - j
   \]
   \[
   \gcd(a_i, a_j) = \gcd(i \cdot m + n + 1 - i, j \cdot m + n + 1 - j)
   \]
   Simplifying, we get:
   \[
   \gcd(i \cdot m + n + 1 - i, j \cdot m + n + 1 - j) = \gcd((j - i) \cdot m, i \cdot m + n + 1 - i)
   \]
   Since \( \gcd((j - i) \cdot m, i \cdot m + n + 1 - i) \) must be 1 for all \( i \neq j \), we need to ensure that \( (j - i) \cdot m \) and \( i \cdot m + n + 1 - i \) are coprime.

3. **Special case when \( n + 1 \) is composite:**
   If \( n + 1 = ab \) where \( 1 < a \leq b \) and \( b > 2 \), then for \( i = a \) and \( j = 2a \), we have:
   \[
   j = 2a < ab \leq n
   \]
   \[
   \gcd(a_i, a_j) = \gcd(a, a(m-1) + ab) = a > 1
   \]
   This implies that if \( n + 1 \) is composite, there exist \( i \) and \( j \) such that \( \gcd(a_i, a_j) > 1 \).

4. **Case when \( n + 1 \) is prime:**
   If \( n + 1 = p \) where \( p \) is a prime number, then \( n = p - 1 \). For \( m = (p - 2)! + 1 \), we have:
   \[
   \gcd(a_i, a_j) = \gcd((j - i)(p - 2)!, i \cdot (p - 2)! + p)
   \]
   Since \( j - i \) divides \( (p - 2)! \) and \( p \) is prime, \( \gcd((j - i)(p - 2)!, i \cdot (p - 2)! + p) = 1 \).

Conclusion:
- If \( n + 1 \) is a composite number, there exist \( i \) and \( j \) such that \( \gcd(a_i, a_j) > 1 \).
- If \( n + 1 \) is a prime number, then \( \gcd(a_i, a_j) = 1 \) for all \( i \neq j \).

The final answer is \( \boxed{ n } \) such that \( n + 1 \) is a prime number.