A hard question in number theory

Try spamming CRT, that is what I would say.

The general way to understand this question is to start with small $n$, start constructing the set, and you gain experience when you fail. After a few failure you should see that making $a-b$ divide both $a$ and $b$ is the easier part, while $a-b$ cannot divide by other elements (let's call it the "exclusive divisibility condition") is the bottleneck.

For brevity I will write down a construction for $n = 5$, but I think you can easily extend it to the general case. Let the set to be constructed called $$\{s_1 = x, s_2 = x+a, s_3 = x+a+b, s_4 = x+a+b+c, s_5 = x+a+b+c+d\}.$$

To satisfy the "exclusive divisibility condition", the heuristics is to make all the differences $$a,b,c,d,a+b,b+c,c+d,a+b+c,b+c+d,a+b+c+d$$ "as pairwise relatively prime as possible". It is impossible to make it perfectly pairwise relatively prime (if you doubt it, you try it). The problem when $n=5$ will be at modulo $2^k$ and $3^k$.

To handle this, we make the following restrictions: $$ \begin{aligned} &a \equiv 2^3 \pmod{2^3}, &a \equiv 3^3 \pmod{3^3};\\ &b \equiv 2^2 \pmod{2^3}, &b \equiv 3^2 \pmod{3^3};\\ &c \equiv 2^1 \pmod{2^3}, &c \equiv 3^1 \pmod{3^3};\\ &d \equiv 2^0 \pmod{2^3}, &d \equiv 3^0 \pmod{3^3}. \end{aligned} $$

This is all prevents the differences to be not pairwise relatively prime. Let $P$ be the set of primes between $5$ and $\binom52$. For any $p \in P$, we also request $$ a \equiv 0 \pmod p, \quad b,c,d \equiv 1 \pmod p. $$

Next we construct $a$. We find $a$ satisfying the restrictions modulo $2^3$, $3^3$, and $P$. The existence is guaranteed by CRT.

Then we construct $b$. Let $D_1$ be the set of prime divisors of $a$. For all $p \in D_1$, we need $b \not\equiv 0 \pmod p$, $b+a \not\equiv 0 \pmod p$, together with the aforementioned restrictions modulo $2^3$, $3^3$, and $P$. The existence is again guaranteed by CRT.

For $c$, we do similar things. $D_2$ be the union of set of prime divisors of $a$, $b$, and $a+b$. we need $c \not\equiv 0 \pmod p$, $c+b \not\equiv 0 \pmod p$, $c+b+a \not\equiv 0 \pmod p$, together with the aforementioned restrictions modulo $2^3$, $3^3$, and $P$. The existence is again guaranteed by CRT.

Do the same for $d$. Let $D$ be the union of the set of prime divisors among all the differences. For $2$ and $3$, you need $x \equiv 0 \pmod{2^33^3}$. For other prime $p \in D$, you decide $x \pmod p$ according to where $p$ came from. For instance, if $p \ | \ b+c$, then you need $x+a \equiv x+a+b+c \equiv 0 \pmod p$. The existence of $x$ is by CRT.

You have to prove that the set we construct is legit. The divisibility condition follows from the construction, and the "exclusive divisibility" is true since all the differences contains a unique prime divisor, and that prime divisor is only divisible by two elements in the set, guaranteed by this construction.