Determine all positive integer numbers $n$ satisfying the following condition:
 the sum of the squares of any $n$ prime numbers greater than $3$ is divisible by $n$.
1. **Definition and Initial Cases**:
   - We define a natural number \( n \) as ***rare*** if the sum of the squares of any \( n \) prime numbers greater than 3 is divisible by \( n \).
   - Clearly, \( n = 1 \) and \( n = 2 \) are rare numbers because:
     - For \( n = 1 \), any single prime number squared is divisible by 1.
     - For \( n = 2 \), the sum of the squares of any two primes greater than 3 is divisible by 2 because the sum of two odd numbers is even.

2. **Assumption for \( n > 2 \)**:
   - Assume \( n > 2 \) is a rare number. Then there exist \( n-1 \) primes \( 7 < p_1 < p_2 < \ldots < p_{n-1} \) such that:
     \[
     n \mid 5^2 + \sum_{i=1}^{n-1} p_i^2 \quad \text{and} \quad n \mid 7^2 + \sum_{i=1}^{n-1} p_i^2
     \]
   - This implies:
     \[
     n \mid 7^2 - 5^2 = 49 - 25 = 24
     \]
   - Therefore, \( n \) must be a divisor of 24.

3. **General Case for \( n \) Distinct Primes**:
   - Assume \( q_1, q_2, \ldots, q_n \) are \( n \) distinct primes greater than 3. Then:
     \[
     S = \sum_{i=1}^n q_i^2
     \]
   - For \( m \in \{3, 8\} \):
     \[
     S \equiv \sum_{i=1}^n q_i^2 \equiv \sum_{i=1}^n 1 = n \pmod{m}
     \]
   - This implies:
     \[
     m \mid S - n
     \]
   - Since 3 and 8 are coprime, by the Chinese Remainder Theorem:
     \[
     24 \mid S - n
     \]

4. **Combining Conditions**:
   - We know \( n \mid S \) (since \( n \) is rare), implying \( n \mid S - n \).
   - Hence, by the conditions \( n \mid 24 \) and \( 24 \mid S - n \), every \( n \) satisfying \( n \mid 24 \) is a rare number.

5. **Conclusion**:
   - The rare numbers are the positive divisors of 24, which are:
     \[
     1, 2, 3, 4, 6, 8, 12, 24
     \]

The final answer is \(\boxed{1, 2, 3, 4, 6, 8, 12, 24}\).