Set of $n$ natural numbers {$a_i$} such that: if $a_j\lt a_k$, then $(a_k-a_j)\mid a_j$

I came across this problem while trying to solve this problem. It did not get much attention, maybe because of poor wording. But it comes down to this:

Prove that for every $n\in N$ there is a set of $n$ different natural numbers $a_i (i=1...n)$, such that if $a_j\lt a_k$, then $(a_k-a_j)\mid a_j$.

Basically, if you pick any two numbers from the set, their difference divides the smaller number.

Constructing such set is trivial for $n=2$ or $n=3$.

For $n=4$, I had to make a few guesses but it was still easy: {8, 9, 10, 12} or {12, 15, 16, 18}.

But I spent the whole hour on $n=5$. The smallest set I could find was {60, 72, 75, 80, 90}, mostly by guessing.

It looks like you can multiply the solution for $n$ with some factor and then squeeze an extra number into the set. But it did not work for me beyond $n=5$ and I gave up on $n=6$ :)

And I still don’t see any pattern. All numbers found so far have a good number of divisors and the solution could have something to do with highly composite numbers. But all my attempts to construct a set in a clever way ($a_i=n!+something(i)$, for example) have failed.

This can be solved by induction:

For $n=2$ you can take set $\{1,2\}$ as the solution.

Now suppose that such set exists for some $n$ and denote that set with $S_n=\{a_i\mid i=1,2,...,n\}$.

Calculate all differencies $d_{jk}$ between all pairs of elements picked from the set $S_n$ and find the least common multiple for all $d_{jk},a_i$:

$$L=LCM\left(\{d_{jk}=a_j-a_k\mid 1\le k <j\le n\}\cup \{a_i\mid i=1,2,...,n\}\right)$$

Check the following set:

$$S_{n+1}=\{L+a_i\mid i=1,2,...,n\} \cup\{L\}$$

This set has $n+1$ elements and all its elements are natural numbers. But also:

$$(L+a_j)-(L+a_k)=a_j-a_k$$

Note that:

$$a_j-a_k\mid L$$ $$a_j-a_k\mid a_k$$

Consequentiallty:

$$(L+a_j)-(L+a_k)\mid L+a_k \qquad(1)$$

If you pick $L$ and $L+a_j$:

$$(L+a_j) - L=a_j\mid L \qquad (2)$$

(1) and (2) complete the induction step.

EDIT:

If you start with:

$$S_2=\{1,2\}$$

and apply the algorithm described above, you get the following sequence of solutions with rapidly growing numbers:

$$S_3=\{2, 3, 4\}$$

$$S_4=\{12, 14, 15, 16\}$$

$$S_5=\{1680, 1692, 1694, 1695, 1696\}$$

$$S_6=\{343319185440, 343319187120, 343319187132, 343319187134, 343319187135, \ 343319187136\}$$

$$S_7=\{118295944058236539689200180716000759951728985298370880, \ 118295944058236539689200180716000759951729328617556320, \ 118295944058236539689200180716000759951729328617558000, \ 118295944058236539689200180716000759951729328617558012, \ 118295944058236539689200180716000759951729328617558014, \ 118295944058236539689200180716000759951729328617558015, \ 118295944058236539689200180716000759951729328617558016\}$$

So the method is far from optimal and finding the "smallest" set for a given $n$ is still a challenge.