Find all squarefree integers whose divisors $d_1 < d_2 < · · · < d_k$ satisfy $d_i − d_{i−1}|n$ for all $2 \leq i \leq k$. Disclaimer:  this problem came from USAMTS: https://www.usamts.org/Tests/Problems_31_3.pdf
The contest has ended., in case there is any doubt. 
Problem: Find all squarefree  integers whose divisors $d_1 < d_2 < · · · < d_k$ satisfy $d_i − d_{i−1}|n$ for all $2 \leq i \leq k$. 
My thinking is if $d_i − d_{i−1}|n$ then $d_i − d_{i−1}$ is in $d_1, d_2,... d_k$. Then $d_2-d_1 < d_2$ is a divisor so $d_2-d_1 = d_1, d_2 = 2d_1$. Obviously $d_1 = 1$ and $d_2 = 2$ since $1 | n$. 
Then $(d_3 - d_2 = d_3 - 2$. if $d_3 - 2 = d_2 $ then $d_3 = 4$. If $d_3 - 2 = d_1 $ then $d_3 = 3$
Powers of 2 seem to work, but not sure how to prove if other cases don't work. 
It seems we can make some decent progress but I am stuck here.. 
 A: We must have $d_1=1.$ So $d_2-1$ must be divisor less than $d_2$ so $d_2-1=1,$ or $d_2=2.$
In general, if $p$ is a prime divisor of $n$ then let $d$ be the previous divisor. Then $p\not\mid d$ and $p-d$ must be a divisor of $n$ and likewise $\frac{n}{d}-\frac{n}{p}=\frac{(p-d)n}{pd}=(p-d)\frac{n}{pd}$ must be a divisor of $n.$ 
But we can show that $\gcd(p-d,pd)=1.$ So if $d\neq p-1$ then $(p-d)\frac{n}{pd}$ is not square free, so it cannot be a divisor of $n.$
So the only way for $p$ to be added is if $p-1$ is also a divisor.
In particular, we must start $1,2,3,\dots,$ and the next cannot be $5$ since $5\neq 3+1.$ So the next must be six, and the next value must be a prime, and the only prime can be $6+1=7.$ 
So we start $1,2,3,6,7,\dots,14,\dots, 21,\dots,42,\dots$ If there were any primes added to this in the $\dots,$ then the smallest such $p$ must have $p-1$ in $7,14,21,42.$ But the only option is $p=43,$ because $7+1,14+1,21+1$ are all non-prime. 
So the sequence must start:
$$1,2,3,6,7,14, 21,42,\dots$$
We can stop at $n=42.$ Or we can continue with $p=43.$ But then we have that there is not prime $p$ so that $p-1=43d_1$ where $d_1\mid 42.$ So there can be no larger $n.$
So the only values are $n=1,2,6=2(2+1),42=6(6+1),1806=42(42+1).$
A: Your progress as far as it goes is spot on. If $d_3=4$, $n$ is not square-free. So you are restricted to looking at $d_3=3$. In this case, $n$ is at least $6$, so $d_4=6$, and $n$ is either $6$ or a multiple of $6$. Any multiple of $6$ that is even or contains a factor of $3$ will not be square-free. So only certain odd multiples of $6$ can be considered.
$d_5-d_4$ must equal $1,2,3,6$, but even numbers and multiples of $3$ are forbidden, so $d_5-d_4=1$ and $d_5=7$, in which case $n$ is either $42$ or a multiple of $42$. In future steps, multiples of $7$ must be avoided to keep the resulting $n$ square-free.
$d_6-d_5$ must equal $1,2,3,6,7,14,21,42$, but even numbers and multiples of $3,7$ are forbidden, so $d_6-d_5=1$ and $d_6=43$, in which case $n$ is either $1806$ or a multiple of $1806$. In future steps, multiples of $43$ must be avoided to keep the resulting $n$ square-free.
$d_7-d_6$ must equal $1,2,3,6,7,14,21,42,43,86,129,258,301,602,903,1806$, but even numbers and multiples of $3,7,43$ are forbidden, so $d_7-d_6=1$ $d_7=1807$, in which case $n$ is $1806\cdot 1807$ or a multiple of $1806\cdot 1807$. Alas, $1807=13\cdot 139$, and $13$ is not in the list of factors, so the trail peters out here.
The possibilities for $n$ are $6,42,1806$
BY EDIT: Oscar Lanzi (in comments) points out that $n=2$ is a solution if the divisors consist entirely of $d_1=1,\ d_2=2$. Apologies for overlooking that as I started with OP's query on how to proceed based on possible values of $d_3$.
