$(CZS 2)$ Let $p$ be a prime odd number. Is it possible to find $p-1$ natural numbers $n + 1, n + 2, . . . , n + p -1$ such that the sum of the squares of these numbers is divisible by the sum of these numbers?
1. **Rewrite the sequence of numbers:**
   Given the sequence \( n+1, n+2, \ldots, n+p-1 \), we can rewrite these numbers in a more symmetric form around their mean. Let \( x = n + \frac{p-1}{2} \). Then the sequence becomes:
   \[
   x - \frac{p-1}{2}, x - \frac{p-3}{2}, \ldots, x + \frac{p-3}{2}, x + \frac{p-1}{2}
   \]

2. **Sum of the numbers:**
   The sum of these \( p-1 \) numbers is:
   \[
   \sum_{k=1}^{p-1} (n+k) = (n+1) + (n+2) + \cdots + (n+p-1)
   \]
   Using the formula for the sum of an arithmetic series, we get:
   \[
   \sum_{k=1}^{p-1} (n+k) = (p-1)n + \sum_{k=1}^{p-1} k = (p-1)n + \frac{(p-1)p}{2}
   \]

3. **Sum of the squares of the numbers:**
   The sum of the squares of these \( p-1 \) numbers is:
   \[
   \sum_{k=1}^{p-1} (n+k)^2 = \sum_{k=1}^{p-1} (n^2 + 2nk + k^2)
   \]
   Simplifying, we get:
   \[
   \sum_{k=1}^{p-1} (n+k)^2 = (p-1)n^2 + 2n \sum_{k=1}^{p-1} k + \sum_{k=1}^{p-1} k^2
   \]
   Using the formulas for the sum of the first \( p-1 \) natural numbers and the sum of the squares of the first \( p-1 \) natural numbers, we get:
   \[
   \sum_{k=1}^{p-1} k = \frac{(p-1)p}{2}
   \]
   \[
   \sum_{k=1}^{p-1} k^2 = \frac{(p-1)p(2p-1)}{6}
   \]
   Therefore:
   \[
   \sum_{k=1}^{p-1} (n+k)^2 = (p-1)n^2 + 2n \cdot \frac{(p-1)p}{2} + \frac{(p-1)p(2p-1)}{6}
   \]
   Simplifying further:
   \[
   \sum_{k=1}^{p-1} (n+k)^2 = (p-1)n^2 + (p-1)np + \frac{(p-1)p(2p-1)}{6}
   \]

4. **Divisibility condition:**
   We need the sum of the squares to be divisible by the sum of the numbers:
   \[
   (p-1)n^2 + (p-1)np + \frac{(p-1)p(2p-1)}{6} \text{ divisible by } (p-1)n + \frac{(p-1)p}{2}
   \]
   Simplifying the divisibility condition:
   \[
   (p-1) \left( n^2 + np + \frac{p(2p-1)}{6(p-1)} \right) \text{ divisible by } (p-1) \left( n + \frac{p}{2} \right)
   \]
   Dividing both sides by \( p-1 \):
   \[
   n^2 + np + \frac{p(2p-1)}{6(p-1)} \text{ divisible by } n + \frac{p}{2}
   \]

5. **Special cases:**
   - For \( p = 3 \):
     \[
     2x + 1 \text{ must divide } 2x^2 + 2x + 1
     \]
     This is not possible as shown in the initial solution.
   
   - For \( p \equiv 5 \pmod{6} \):
     \[
     x = \frac{p-3}{3} \cdot \frac{p-2}{2} + \frac{p-1}{2}
     \]
     This provides a solution.

   - For \( p \equiv 1 \pmod{6} \):
     It is clear that there are no solutions because the right-hand side is not a multiple of \( \frac{p-1}{2} \).

Conclusion:
The problem has solutions only if \( p \equiv 5 \pmod{6} \).

The final answer is \( \boxed{ p \equiv 5 \pmod{6} } \).