A positive integer $n > 1$ is juicy if its divisors $d_1 < d_2 < \dots < d_k$ satisfy $d_i - d_{i-1} \mid n$ for all $2 \leq i \leq k$. Find all squarefree juicy integers.
1. **Identify the form of $n$:**
   Given that $n$ is a positive integer greater than 1 and is juicy, we need to find all squarefree juicy integers. A number $n$ is juicy if its divisors $d_1 < d_2 < \dots < d_k$ satisfy $d_i - d_{i-1} \mid n$ for all $2 \leq i \leq k$.

2. **Consider the largest and second largest divisors:**
   The largest divisor of $n$ is $n$ itself, and the second largest divisor is $\frac{n}{k}$ for some integer $k$. We must have:
   \[
   n - \frac{n}{k} \mid n
   \]
   Simplifying, we get:
   \[
   \frac{n}{n - \frac{n}{k}} = \frac{1}{1 - \frac{1}{k}} = \frac{k}{k-1} \in \mathbb{Z}
   \]
   The only integer value for which this holds true is $k = 2$. Thus, $n$ must be of the form $2p_1 p_2 \dots p_m$, where $p_i$ are distinct primes.

3. **Analyze the divisors:**
   For each divisor $d_i$, we have $d_i = \frac{n}{d_{k-i+1}}$. Therefore:
   \[
   d_{i+1} - d_i \mid n \implies \frac{n}{d_{k-i}} - \frac{n}{d_{k-i+1}} \mid n
   \]
   Simplifying, we get:
   \[
   \frac{1}{\frac{1}{d_{k-i}} - \frac{1}{d_{k-i+1}}} \in \mathbb{Z} \implies \frac{d_{k-i+1} d_{k-i}}{d_{k-i+1} - d_{k-i}} \in \mathbb{Z}
   \]
   Since $k - i + 1 = (k-i) + 1$, we must have:
   \[
   \frac{d_i d_{i-1}}{d_i - d_{i-1}} \in \mathbb{Z}
   \]
   for all divisors $d_i$. Thus, $d_i d_{i-1} \equiv 0 \pmod{d_i - d_{i-1}}$.

4. **Constructing juicy numbers:**
   - The first two factors of $n$ are $n$ and $\frac{n}{2}$.
   - If the next factor is of the form $\frac{n}{k}$, then $\frac{2k}{k-2}$ must be an integer. This holds for $k = 3$ and $k = 4$. Since $n$ is squarefree, only $k = 3$ works, implying $n = 6$.
   - The next factor can be $\frac{n}{5}$ or $\frac{n}{6}$. If $\frac{n}{5}$ is a factor, then $5 \mid n$, which contradicts $n$ being squarefree. Thus, $\frac{n}{6}$ is the next factor, implying $n = 2$ and $n = 6$ are possible solutions.
   - For $n$ with three prime factors, $n = 2 \cdot 3 \cdot 7 = 42$.
   - For $n$ with four prime factors, $n = 42p$. Testing primes, we find $p = 43$ works, giving $n = 42 \cdot 43 = 1806$.

5. **Verification and conclusion:**
   - For $n = 2$, $6$, $42$, and $1806$, we verify that they satisfy the juicy condition.
   - For $n = 1806p$, $p - 1 \mid 1806$. Testing possible values, none are prime, confirming no other solutions.

The final answer is $\boxed{2, 6, 42, 1806}$