Call a natural number $n$ [i]good[/i] if for any natural divisor $a$ of $n$, we have that $a+1$ is also divisor of $n+1$. Find all good natural numbers.

[i]S. Berlov[/i]
1. **Base Case: \( n = 1 \)**
   - The divisors of \( n = 1 \) are \( 1 \).
   - For \( a = 1 \), \( a + 1 = 2 \).
   - We need to check if \( 2 \) is a divisor of \( n + 1 = 2 \). Indeed, \( 2 \) is a divisor of \( 2 \).
   - Therefore, \( n = 1 \) is a good number.

2. **Case: \( n \) is an odd prime**
   - Let \( n = p \) where \( p \) is an odd prime.
   - The divisors of \( p \) are \( 1 \) and \( p \).
   - For \( a = 1 \), \( a + 1 = 2 \). We need to check if \( 2 \) is a divisor of \( p + 1 \). Since \( p \) is odd, \( p + 1 \) is even, and thus \( 2 \) is a divisor of \( p + 1 \).
   - For \( a = p \), \( a + 1 = p + 1 \). We need to check if \( p + 1 \) is a divisor of \( p + 1 \). This is trivially true.
   - Therefore, any odd prime \( p \) is a good number.

3. **General Case: \( n \) is composite**
   - Assume \( n = ab \) for some \( a, b > 1 \) with \( a \ge b \).
   - Since \( a \mid n \), we have \( a \mid n - a \).
   - We need \( a + 1 \mid n + 1 \). This implies \( a + 1 \mid (n + 1) - (n - a) = a + 1 \).
   - Therefore, \( a(a + 1) \mid n - a \).
   - This implies \( a(a + 1) \le n - a \le n \).
   - Since \( n = ab \), we have \( a(a + 1) \le ab \).
   - Dividing both sides by \( a \), we get \( a + 1 \le b \).
   - Since \( a \ge b \), we have \( a + 1 \le a \), which is a contradiction.
   - Therefore, no composite number \( n \) can be a good number.

Conclusion:
- The only good natural numbers are \( 1 \) and all odd primes.

The final answer is \( \boxed{ 1 } \) and all odd primes.