We call a number [i]perfect[/i] if the sum of its positive integer divisors(including $1$ and $n$) equals $2n$. Determine all [i]perfect[/i] numbers $n$ for which $n-1$ and $n+1$ are prime numbers.
1. **Definition and Initial Assumption**:
   We start by defining a perfect number \( n \) as a number for which the sum of its positive divisors (including \( 1 \) and \( n \)) equals \( 2n \). We need to determine all perfect numbers \( n \) such that both \( n-1 \) and \( n+1 \) are prime numbers.

2. **Verification for \( n = 6 \)**:
   The first perfect number is \( n = 6 \). We check if \( n-1 \) and \( n+1 \) are prime:
   \[
   n-1 = 5 \quad \text{(prime)}, \quad n+1 = 7 \quad \text{(prime)}
   \]
   Thus, \( n = 6 \) satisfies the conditions.

3. **Assumption for \( n > 6 \)**:
   Assume \( n > 6 \) and suppose \( n \) is a perfect number such that \( n-1 \) and \( n+1 \) are prime.

4. **Divisibility by 3**:
   Since \( n-1 \) and \( n+1 \) are prime, they cannot be multiples of 3. Therefore, \( n \) must be a multiple of 3. This is because if \( n \) were not a multiple of 3, one of \( n-1 \), \( n \), or \( n+1 \) would be a multiple of 3, contradicting the primality of \( n-1 \) and \( n+1 \).

5. **Evenness of \( n \)**:
   Since \( n-1 \) and \( n+1 \) are both odd (primes greater than 2), \( n \) must be even. Therefore, \( n \) is a multiple of 6 (since it is both even and a multiple of 3).

6. **Sum of Divisors**:
   For \( n \) to be perfect, the sum of its divisors must equal \( 2n \). Consider the divisors of \( n \):
   \[
   n, \quad \frac{n}{2}, \quad \frac{n}{3}, \quad \frac{n}{6}, \quad 1
   \]
   These are distinct divisors of \( n \) since \( n > 6 \). The sum of these divisors is:
   \[
   n + \frac{n}{2} + \frac{n}{3} + \frac{n}{6} + 1
   \]
   Simplifying the sum:
   \[
   n + \frac{n}{2} + \frac{n}{3} + \frac{n}{6} + 1 = n + \frac{3n}{6} + \frac{2n}{6} + \frac{n}{6} + 1 = n + \frac{6n}{6} + 1 = n + n + 1 = 2n + 1
   \]
   This sum is strictly greater than \( 2n \), which contradicts the definition of a perfect number.

7. **Conclusion**:
   Since \( n > 6 \) leads to a contradiction, the only perfect number \( n \) for which \( n-1 \) and \( n+1 \) are prime is \( n = 6 \).

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