Find all the positive integers $n$ such that:
$(1)$ $n$ has at least $4$ positive divisors.
$(2)$ if all positive divisors of $n$ are $d_1,d_2,\cdots ,d_k,$ then $d_2-d_1,d_3-d_2,\cdots ,d_k-d_{k-1}$ form a geometric sequence.
1. **Express \( n \) in terms of its prime factorization:**
   Let \( n = \prod_{i=1}^r p_i^{\alpha_i} \), where \( p_i \) are distinct primes and \( \alpha_i \geq 1 \). We are given that \( n \) has at least 4 positive divisors, so \( k \geq 4 \).

2. **Assume \( r \geq 2 \) and derive a contradiction:**
   Suppose \( r \geq 2 \). The divisors of \( n \) are \( d_1, d_2, \ldots, d_k \) in increasing order. The differences \( d_2 - d_1, d_3 - d_2, \ldots, d_k - d_{k-1} \) form a geometric sequence with common ratio \( x \).

3. **Analyze the differences:**
   - The largest difference is \( d_k - d_{k-1} = n - \frac{n}{p_1} = n \left(1 - \frac{1}{p_1}\right) \).
   - The smallest difference is \( d_2 - d_1 = p_1 - 1 \).

4. **Set up the geometric sequence:**
   Let the common ratio be \( x \). Then:
   \[
   \frac{n \left(1 - \frac{1}{p_1}\right)}{p_1 - 1} = x^{k-2}
   \]
   Simplifying, we get:
   \[
   \frac{n - \frac{n}{p_1}}{p_1 - 1} = x^{k-2}
   \]
   \[
   \frac{n \left(1 - \frac{1}{p_1}\right)}{p_1 - 1} = x^{k-2}
   \]
   \[
   \frac{n \left(\frac{p_1 - 1}{p_1}\right)}{p_1 - 1} = x^{k-2}
   \]
   \[
   \frac{n}{p_1} = x^{k-2}
   \]

5. **Compare exponents:**
   Since \( x \) is a rational number, let \( x = \prod_{i=1}^r p_i^{\beta_i} \). Then:
   \[
   \frac{n}{p_1} = \prod_{i=1}^r p_i^{\alpha_i - \delta_{i1}} = \prod_{i=1}^r p_i^{\beta_i (k-2)}
   \]
   Comparing the exponents, we get:
   \[
   \alpha_i - \delta_{i1} = \beta_i (k-2)
   \]
   For \( i \geq 2 \), \( \alpha_i = \beta_i (k-2) \). For \( i = 1 \), \( \alpha_1 - 1 = \beta_1 (k-2) \).

6. **Derive a contradiction:**
   Since \( k \geq 4 \), \( k-2 \geq 2 \). Thus, \( k-2 \) must divide \( \alpha_i \) for \( i \geq 2 \) and \( \alpha_1 - 1 \). This implies \( k-1 \leq \alpha_i \) for \( i \geq 2 \). But:
   \[
   k-1 = \prod_{i=1}^r (\alpha_i + 1) - 1 \geq 2 (\alpha_i + 1) - 1 = 2 \alpha_i + 1 > \alpha_i
   \]
   This is a contradiction. Therefore, \( r \geq 2 \) is not possible.

7. **Conclude \( r = 1 \):**
   Since \( r \geq 2 \) leads to a contradiction, we must have \( r = 1 \). Thus, \( n = p^\alpha \) for some prime \( p \) and integer \( \alpha \geq 3 \).

8. **Verify the solution:**
   If \( n = p^\alpha \) with \( \alpha \geq 3 \), the divisors of \( n \) are \( 1, p, p^2, \ldots, p^\alpha \). The differences \( p-1, p(p-1), p^2(p-1), \ldots \) form a geometric sequence with ratio \( p \).

Thus, the set of positive integers \( n \) that satisfy the given property is:
\[
S = \{ p^\alpha \mid p \text{ prime}, \alpha \geq 3, \alpha \in \mathbb{N} \}
\]

The final answer is \( \boxed{ S = \{ p^\alpha \mid p \text{ prime}, \alpha \geq 3, \alpha \in \mathbb{N} \} } \).