Find all infinite arithmetic progressions formed with positive integers such that there exists a number $N \in \mathbb{N}$, such that for any prime $p$, $p > N$,
the $p$-th term of the progression is also prime.
1. **Restate the problem in terms of arithmetic progression:**
   We need to find all infinite arithmetic progressions of positive integers such that there exists a number \( N \in \mathbb{N} \) such that for any prime \( p \) with \( p > N \), the \( p \)-th term of the progression is also prime.

2. **Express the arithmetic progression:**
   An arithmetic progression can be written as \( a_n = a + kn \) where \( a \) and \( k \) are integers, and \( n \) is a positive integer.

3. **Reformulate the problem:**
   We need to determine all pairs of integers \( (a, k) \) such that for all sufficiently large prime numbers \( p > N \), the term \( a + kp \) is also a prime number.

4. **Consider the case \( k = 0 \):**
   If \( k = 0 \), the sequence is constant, i.e., \( a_n = a \). For the sequence to satisfy the condition, \( a \) must be a prime number. Thus, \( a = P \) where \( P \) is a prime number.

5. **Consider the case \( k > 0 \):**
   - If \( k > 1 \), then \( a \neq 0 \). Take a large prime \( Q \) such that \( \gcd(a, k) = 1 \). By Dirichlet's theorem on arithmetic progressions, there are infinitely many primes \( P \) such that:
     \[
     P \equiv -\frac{a}{k} \pmod{Q}
     \]
     This implies \( a + kP \equiv 0 \pmod{Q} \), meaning \( a + kP \) is a multiple of \( Q \). Since \( a + kP \) must be prime and \( Q \) is large, \( a + kP \) must equal \( Q \). This is a contradiction because there are infinitely many such \( P \), but \( Q \) is fixed.

6. **Conclude \( k = 1 \):**
   - If \( k = 1 \), the sequence is \( a_n = a + n \). For \( a + p \) to be prime for all sufficiently large primes \( p \), \( a \) must be 0. Otherwise, there would be a large prime \( p \) such that \( a + p \) is not prime.

7. **Conclusion:**
   The only arithmetic progressions that satisfy the given condition are:
   - \( a_n = P \) for some prime \( P \)
   - \( a_n = n \)

The final answer is \( \boxed{ a_n = P } \) for some prime \( P \) and \( a_n = n \).