2. 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.
Given the conditions $k \geqslant 4$, and 
\[
\frac{d_{3}-d_{2}}{d_{2}-d_{1}}=\frac{d_{k}-d_{k-1}}{d_{k-1}-d_{k-2}},
\]
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 equation, we get
\[
\frac{d_{3}-d_{2}}{d_{2}-1}=\frac{n-\frac{n}{d_{2}}}{\frac{n}{d_{2}}-\frac{n}{d_{3}}}=\frac{d_{2} d_{3}-d_{3}}{d_{3}-d_{2}} \Rightarrow\left(d_{3}-d_{2}\right)^{2}=d_{3}\left(d_{2}-1\right)^{2}.
\]

Thus, $d_{3}$ 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 $d_{3}=p^{2}$. Therefore, the sequence $d_{2}-d_{1}, d_{3}-d_{2}, \cdots, d_{k}-d_{k-1}$ is $p-1, p^{2}-p, p^{3}-p^{2}$, $\cdots, p^{k-1}-p^{k-2}$, and the corresponding $n=p^{k-1}$.

In summary, the $n$ that satisfies the conditions are all numbers of the form $p^{\alpha}$, where $p$ is a prime number, and the integer $\alpha \geqslant 3$.