Let $N$ be a positive integer; a divisor of $N$ is called [i]common[/i] if it's great than $1$ and different of $N$. A positive integer is called [i]special[/i] if it has, at least, two common divisors and it is multiple of all possible differences between any two of their common divisors. 
Find all special integers.
1. **Initial Considerations**:
   - A positive integer \( N \) is called *special* if it has at least two common divisors and is a multiple of all possible differences between any two of its common divisors.
   - A common divisor of \( N \) is a divisor greater than 1 and different from \( N \).

2. **Case 1: \( N \) is odd**:
   - If \( N \) is odd, all common divisors of \( N \) are also odd.
   - The difference between any two odd numbers is even.
   - Therefore, \( N \) must be even to be a multiple of 2 (since 2 divides the difference of any two odd common divisors).
   - This leads to a contradiction, so \( N \) must be even.

3. **Case 2: \( N \) is a power of 2**:
   - Let \( N = 2^k \).
   - The common divisors of \( N \) are \( 2, 4, 8, \ldots, 2^{k-1} \).
   - For \( k \geq 3 \), the difference \( 8 - 2 = 6 \) must divide \( N \).
   - Therefore, \( N \leq 8 \).
   - Checking \( N = 8 \), the common divisors are \( 2, 4 \), and the difference \( 4 - 2 = 2 \) divides 8.
   - Thus, \( N = 8 \) is a solution.

4. **Case 3: \( N \) has a prime divisor greater than 2**:
   - Let \( p \) be the smallest prime divisor of \( N \) greater than 2.
   - Then \( p - 2 \) must divide \( N \).
   - If \( p > 3 \), this is impossible because \( p - 2 \) is not a divisor of \( N \).
   - Therefore, \( p = 3 \), and \( N \) must be of the form \( 6k \).

5. **Checking \( N = 6k \)**:
   - For \( k = 1 \), \( N = 6 \):
     - The common divisors are \( 2, 3 \), and the difference \( 3 - 2 = 1 \) divides 6.
     - Thus, \( N = 6 \) is a solution.
   - For \( k = 2 \), \( N = 12 \):
     - The common divisors are \( 2, 3, 4, 6 \), and the differences are \( 1, 2, 3, 4 \), all of which divide 12.
     - Thus, \( N = 12 \) is a solution.
   - For \( k = 3, 4, 5, 6, 7 \), checking each case does not yield new solutions.
   - For \( k > 7 \):
     - We need to check if \( 3k - 2 \) and \( 2k - 3 \) divide \( N \).
     - The greatest common divisor \( \gcd(2k - 3, 3k - 2) \leq 5 \).
     - Therefore, \( (2k - 3)(3k - 2) \) must divide \( 30k \).
     - Solving \( 6k^2 - 13k + 6 \leq 30k \) gives \( 6k^2 - 43k + 6 \leq 0 \), implying \( k < 8 \).

6. **Conclusion**:
   - The only solutions are \( N = 6, 8, 12 \).

The final answer is \( \boxed{6, 8, 12} \).