Find the largest positive integer $n$ such that for each prime $p$ with $2<p<n$ the difference $n-p$ is also prime.
1. We need to find the largest positive integer \( n \) such that for each prime \( p \) with \( 2 < p < n \), the difference \( n - p \) is also prime.
2. Let's assume \( n > 10 \). Consider the primes \( p = 3, 5, 7 \) which are all less than \( n \).
3. Then, the differences \( n - 3 \), \( n - 5 \), and \( n - 7 \) must all be primes.
4. Notice that any integer greater than 3 is either of the form \( 3k \), \( 3k+1 \), or \( 3k+2 \) for some integer \( k \).
5. Since \( n \) is greater than 10, at least one of \( n-3 \), \( n-5 \), or \( n-7 \) must be divisible by 3 (because they form a complete residue system modulo 3).
6. If any of these differences is divisible by 3 and greater than 3, it cannot be prime (since the only prime divisible by 3 is 3 itself).
7. This leads to a contradiction, implying that \( n \leq 10 \).

8. Now, we need to check if \( n = 10 \) satisfies the condition:
   - For \( p = 3 \), \( n - p = 10 - 3 = 7 \), which is prime.
   - For \( p = 5 \), \( n - p = 10 - 5 = 5 \), which is prime.
   - For \( p = 7 \), \( n - p = 10 - 7 = 3 \), which is prime.

9. Therefore, \( n = 10 \) satisfies the condition that for each prime \( p \) with \( 2 < p < n \), the difference \( n - p \) is also prime.

The final answer is \( \boxed{10} \).