Determine all natural numbers$ n> 1$ with the property: 
For each divisor $d> 1$  of number $n$, then $d - 1$ is a divisor of $n - 1$.
1. **Identify the problem statement and initial conditions:**
   We need to determine all natural numbers \( n > 1 \) such that for each divisor \( d > 1 \) of \( n \), \( d - 1 \) is a divisor of \( n - 1 \).

2. **Consider the case when \( n \) is a prime number:**
   - If \( n \) is a prime number, then its only divisors greater than 1 are \( n \) itself.
   - For \( d = n \), we have \( d - 1 = n - 1 \).
   - Clearly, \( n - 1 \mid n - 1 \), so the condition is satisfied for prime numbers.

3. **Consider the case when \( n \) is a composite number:**
   - Let \( n \) be a composite number and \( d \) be a divisor of \( n \) such that \( 1 < d < n \).
   - We need to show that \( d - 1 \mid n - 1 \).

4. **Analyze the given condition:**
   - Given \( d \mid n \) and \( d > 1 \), we need \( d - 1 \mid n - 1 \).
   - Consider the expression \( \frac{n - 1}{d - 1} - \frac{n}{d} \):
     \[
     \frac{n - 1}{d - 1} - \frac{n}{d} = \frac{n - d}{d(d - 1)}
     \]
   - For this to be an integer, \( \frac{n - d}{d(d - 1)} \) must be an integer, implying:
     \[
     \frac{n - d}{d(d - 1)} \geq 1 \implies n - d \geq d(d - 1) \implies n \geq d^2
     \]
   - Therefore, \( d \leq \sqrt{n} \).

5. **Consider the least prime divisor \( p \) of \( n \):**
   - Let \( p \) be the least prime divisor of \( n \).
   - Since \( p \leq \sqrt{n} \), we have \( \frac{n}{p} \geq \sqrt{n} \).
   - If \( \frac{n}{p} \) is also a divisor of \( n \), then \( n = p^2 \) is a possibility.

6. **Verify the condition for \( n = p^2 \):**
   - If \( n = p^2 \), then \( d = p \) is a divisor of \( n \).
   - We need to check if \( p - 1 \mid p^2 - 1 \):
     \[
     p^2 - 1 = (p - 1)(p + 1)
     \]
   - Clearly, \( p - 1 \mid (p - 1)(p + 1) \), so the condition is satisfied.

7. **Summarize the findings:**
   - The condition \( d - 1 \mid n - 1 \) is satisfied if \( n \) is either a prime number or a square of a prime number.

The final answer is \( \boxed{ n } \) is either a prime number or a square of a prime number.