Let $n$ be an integer greater than 1, and its all positive divisors are $d_{1}, d_{2}, \cdots, d_{k}$, where $1=d_{1}<d_{2}<\cdots<d_{k}=n$. It is known that $d_{2}-d_{1}, d_{3}-d_{2}, \cdots, d_{k}-d_{k-1}$ are all the positive divisors of another positive integer (all distinct but not necessarily in increasing order). Find all possible values of $n$.
Let $d_{i}-d_{i-1}=\delta_{i}(i=2,3, \cdots, n)$. It is easy to see that the number with the set of positive divisors $D=\left\{\delta_{2}, \delta_{3}, \cdots, \delta_{k}\right\}$ is the largest element in $D$. Since $1 \in D$, we know that the set $\left\{d_{1}, d_{2}, \cdots, d_{k}\right\}$ contains both odd and even numbers, hence $n$ is even. Therefore, $d_{k}=n, d_{k-1}=\frac{n}{2}$, which means $\delta_{k}=\frac{n}{2}$, and the other elements $\delta_{i}<d_{i} \leqslant \frac{n}{2}$ in $D$, so $D$ is the set of positive divisors of $\frac{n}{2}$. Thus, the number of positive divisors of $n$ is one more than that of $\frac{n}{2}$.

Let $n=2^{s} \cdot q\left(s \in \mathbf{N}^{*}, q\right.$ be an odd number $)$, and denote the number of positive divisors of a positive integer $m$ as $d(m)$, then $d(n)=(s+1) d(q), d\left(\frac{n}{2}\right)=s d(q)$, so $d(q)=1$, which implies $q=1$.
Therefore, $n=2^{s}$, at this point the set of positive divisors of $n$ is $\left\{1,2, \cdots, 2^{s}\right\}, D=\left\{1,2, \cdots, 2^{s-1}\right\}$, which meets the conditions. Hence, all $n$ that satisfy the conditions are $n=2^{s}\left(s \in \mathbf{N}^{*}\right)$.