Problem: Prove that for every positive integer $n$, there exists a set $S$ of $n$ positive integers
such that for any two distinct $a,b \in S$, $a-b$ divides $a$ and $b$
but none of the other elements of $S$.

Solution: The idea is to look for a
sequence $d_1, \dots, d_{n-1}$ of ``differences''
such that the following two conditions hold.
Let $s_i = d_1 + \dots + d_{i-1}$,
and $t_{i,j} = d_i + \dots + d_{j-1}$ for $i \le j$.
\begin{enumerate}
\item[(i)] No two of the $t_{i,j}$ divide each other.
\item[(ii)] There exists an integer $a$
   satisfying the CRT equivalences
   \[ a \equiv -s_i \pmod{t_{i,j}} \quad \forall i \le j \]
\end{enumerate}
Then the sequence $a+s_1$, $a+s_2$, \dots, $a+s_n$ will work.
For example, when $n=3$ we can take $(d_1,d_2)=(2,3)$ giving
\[ 10 \overbrace{\underbrace{\qquad}_2 12 \underbrace{\qquad}_3}^5 15 \]
because the only conditions we need satisfy are
\begin{align*}
  a &\equiv 0 \pmod 2 \\
  a &\equiv 0 \pmod 5 \\
  a &\equiv -2 \pmod 3.
\end{align*}

But with this setup we can just construct the $d_i$ inductively.
To go from $n$ to $n+1$, take a $d_1, \dots, d_{n-1}$
and let $p$ be a prime
not dividing any of the $d_i$.
Moreover, let $M$ be a multiple of $\prod_{i \le j} t_{i,j}$ coprime to $p$.
Then we claim that $d_1M, d_2M, \dots, d_{n-1}M, p$
is such a difference sequence.
For example, the previous example extends as follows with $M = 300$ and $p=7$.
\[
   a \overbrace{\underbrace{\qquad}_{600} b
   \overbrace{\underbrace{\qquad}_{900} c
   \underbrace{\qquad}_{7}}^{907}}^{1507} d
\]
The new numbers $p$, $p+Mt_{n-1,n}$, $p+Mt_{n-2,n}$, \dots\
are all relatively prime to everything else.
Hence (i) still holds.
To see that (ii) still holds,
just note that we can still get a family of solutions
for the first $n$ terms, and then the last $(n+1)$st term
can be made to work by Chinese Remainder Theorem
since all the new $p+Mt_{n-2,n}$ are coprime to everything.