II. (This question is worth 40 points) Find all positive integers $n$ that satisfy the following conditions:
(1) $n$ has at least 4 positive divisors:
(2) If $d_{1}<d_{2}<\cdots<d_{k}$ are all the positive divisors of $n$, then $d_{2}-d_{1}, d_{3}-d_{2}, \cdots, d_{k}-d_{k-1}$ form a geometric sequence.
Solution: From the conditions, we know that $k \geq 4$, and 
$$
\frac{d_{3}-d_{2}}{d_{2}-d_{1}}=\frac{d_{k}-d_{k-1}}{d_{k-1}-d_{k-2}}.
$$
10 points
It is easy to see that $d_{1}=1, d_{k}=n, d_{k-1}=\frac{n}{d_{2}}, d_{k-2}=\frac{n}{d_{3}}$, substituting these into the above equation, we get
$$
\frac{d_{3}-d_{2}}{d_{2}-1}=\frac{n-\frac{n}{d_{2}}}{\frac{n}{d_{2}}-\frac{n}{d_{3}}},
$$

Simplifying, we get
$$
\left(d_{3}-d_{2}\right)^{2}=\left(d_{2}-1\right)^{2} d_{3} \text {. }
$$

From this, we know that $d_{3}$ is a perfect square. Since $d_{2}=p$ is the smallest prime factor of $n$, and $d_{3}$ is a square number, it can only be that $d_{3}=p^{2}$. 30 points
Thus, the sequence $d_{2}-d_{1}, d_{3}-d_{2}, \cdots, d_{k}-d_{k-1}$ is $p-1, p^{2}-p, p^{3}-p^{2}, \cdots, p^{k-1}-p^{k-2}$, which means $d_{1}, d_{2}, d_{3}, \cdots, d_{k}$ are $1, p, p^{2}, \cdots, p^{k-1}$, and the corresponding $n$ is $p^{k-1}$.
In summary, the $n$ that satisfies the conditions are all numbers of the form $p^{a}$, where $p$ is a prime number, and the integer $a \geq 3$.