Find all odd positive integers $ n > 1$ such that if $ a$ and $ b$ are relatively prime divisors of $ n$, then $ a\plus{}b\minus{}1$ divides $ n$.
1. **Express \( n \) in terms of its prime factors:**
   Let \( n = \prod_{k=1}^{t} p_k^{x_k} \), where \( p_k \) are distinct prime factors of \( n \) and \( x_k \) are their respective powers.

2. **Choose \( a \) and \( b \) as relatively prime divisors of \( n \):**
   Let \( a = p_1^{x_1} \) be the smallest prime power factor of \( n \). Then, choose \( b = \frac{n}{a} \).

3. **Set up the divisibility condition:**
   Since \( a \) and \( b \) are relatively prime, we have:
   \[
   a + b - 1 \mid n
   \]
   Substituting \( b = \frac{n}{a} \), we get:
   \[
   a + \frac{n}{a} - 1 \mid n
   \]
   This implies:
   \[
   a^2 - a + n \mid an
   \]

4. **Simplify the divisibility condition:**
   Since \( a^2 - a + n \mid an \), it follows that:
   \[
   a^2 - a + n \mid a^3 - a^2
   \]
   Therefore:
   \[
   a^2 - a + n \leq a^3 - a^2
   \]
   Simplifying, we get:
   \[
   n \leq a(a-1)^2 < a^3
   \]

5. **Analyze the number of prime factors \( t \):**
   Since \( n > a^t \) and \( n < a^3 \), it follows that:
   \[
   t < 3 \implies t \in \{1, 2\}
   \]

6. **Case \( t = 1 \):**
   If \( t = 1 \), then \( n = p^x \) for some odd prime \( p \) and positive integer \( x \). This satisfies the condition.

7. **Case \( t = 2 \):**
   If \( t = 2 \), then \( n = p^x q^y \) for distinct primes \( p \) and \( q \). Assume \( p < q \).

8. **Check the condition for \( t = 2 \):**
   - Choose \( a = p \) and \( b = q \). Then:
     \[
     p + q - 1 \mid n
     \]
     This implies:
     \[
     q \mid p - 1
     \]
     which is impossible since \( p < q \).

   - Therefore, \( p + q - 1 = p^z \) for some integer \( z \). This implies:
     \[
     q = p^z - p + 1
     \]

9. **Further analysis for \( t = 2 \):**
   - Choose \( a = p^2 \) and \( b = q \). Then:
     \[
     p^2 + q - 1 \mid n
     \]
     This implies:
     \[
     p^z + p^2 - p \mid n
     \]
     Therefore:
     \[
     p^{z-1} + p - 1 \mid n
     \]
     Since \( \gcd(p^{z-1} + p - 1, p^x) = 1 \), it follows that:
     \[
     p^{z-1} + p - 1 = q^\gamma = (p^z - p + 1)^\gamma
     \]

10. **Check for contradictions:**
    - Clearly, \( z > 1 \) (otherwise \( q = 1 \), which is impossible).
    - Checking modulo \( p \) gives:
      \[
      -1 \equiv 1 \pmod{p} \implies p \mid 2
      \]
      which contradicts the fact that \( p \) is odd.

11. **Conclusion:**
    The only possible solution is when \( t = 1 \), i.e., \( n = p^m \) for some odd prime \( p \) and positive integer \( m \).

The final answer is \( \boxed{p^m \mid p \text{ is odd prime}, m \in \mathbb{N}} \).