Find all positive integers n with the following property: For every positive divisor $d$ of $n$, $d+1$ divides $n+1$.
To find all positive integers \( n \) such that for every positive divisor \( d \) of \( n \), \( d+1 \) divides \( n+1 \), we proceed as follows:

1. **Assume \( n \) is composite**:
   Let \( n = pk \) where \( p \) is the smallest prime divisor of \( n \) and \( k \) is an integer. We need to show that \( k+1 \mid pk+1 \).

2. **Use the given property**:
   Since \( k \) is a divisor of \( n \), by the given property, \( k+1 \mid n+1 \). Therefore, \( k+1 \mid pk+1 \).

3. **Apply the Euclidean algorithm**:
   To simplify \( k+1 \mid pk+1 \), we use the Euclidean algorithm:
   \[
   \gcd(k+1, pk+1) = \gcd(k+1, (pk+1) - p(k+1)) = \gcd(k+1, pk+1 - pk - p) = \gcd(k+1, p-1)
   \]

4. **Analyze the gcd**:
   Since \( p \) is the smallest prime divisor of \( n \), \( k \geq p \). Therefore, \( k+1 \geq p+1 \). This implies:
   \[
   \gcd(k+1, p-1) \leq p-1 < k+1
   \]
   Hence, \( k+1 \) cannot be a divisor of \( pk+1 \) unless \( k+1 = p-1 \), which is a contradiction because \( k+1 > p-1 \).

5. **Conclude that \( n \) must be prime**:
   Since \( n \) cannot be composite, \( n \) must be a prime number.

6. **Verify that all odd primes work**:
   For any odd prime \( p \), the divisors of \( p \) are \( 1 \) and \( p \). We need to check:
   - For \( d = 1 \), \( d+1 = 2 \) divides \( p+1 \) since \( p \) is odd.
   - For \( d = p \), \( d+1 = p+1 \) divides \( p+1 \).

   Therefore, all odd primes satisfy the given property.

The final answer is \( \boxed{ n } \) must be an odd prime.