We will prove by induction on $n$, that we can find such a set $S_n$ with non-negative elements. For $n=2$ take $S_n=\{0,1\}$ Now suppose that for some $n\geq 2$, the desired set $S_n$ of $n$ non-negative integers exists. Let $L$ be the least sommon multiple of those numbers $(a-b)^2$ and $ab$ that are non-zero, with $(a,b)$ ranging over pairs of distince elements from $S_n$. Now define: $S_{n+1}=\{L+a: a\in S_n\}\cup \{0\}$ Then $S_{n+1}$ consists of $n+1$ non-negative integers, since $L>0$. If $i,j\in S_{n+1}$ and either $i$ or $j$ is zero, then $(i-j)^2$ divids $ij$. If $L+a, L+b\in S_{n+1}$ with $a,b$, distinct elements of $S_n$, then $(L+a)(L+b)\equiv ab\equiv 0 \pmod{(a-b)^2}$ so $[(L+a)-(L+b)]^2$ divides $(L+a)(L+b)$ thus completeing the induction step $\mathbb{QED}$

We define $ a!^n$ to be $ (a!)!^{n - 1}$. For example, $ 3!^3 = (3!)!^2 = 6!^2 = 720!$. Now, let $ g = \gcd (a,b)$ where $ a$ and $ b$ are part of $ S$, so $ a = a_1g$ and $ b = b_1g$. We then have that $ \frac {ab}{(a - b)^2} = \frac {a_1b_1}{(a_1 - b_1)^2}$. Yet, $ \gcd (a_1 - b_1, a_1) = 1$ and $ \gcd (a_1 - b_1, b_1) = 1$, so $ a_1 - b_1$ can only divide $ a_1b_1$ if $ a_1 - b_1 = 1$. This means that we must have $ (a - b)|a, b$. We now consider the numbers:\[ n!^{2n}, n!^{2n} + n!^{2n - 2}, n!^{2n} + n!^{2n - 2} + n!^{2n - 4}, \cdots , n!^{2n} + n!^{2n - 2} + n!^{2n - 4} + \cdots + n!^2\]We will show that these numbers work. Define $ a_i$ to be $ n!^{2n} + n!^{2n - 2} + \cdots + n!^{2n - 2i + 2}$. We notice that $ a_i - a_j$ for $ i > j$ is equal to \[ n!^{2n - 2i + 2} + n!^{2n - 2i + 4} + \cdots + n!^{2n - 2j} < n\cdot n!^{2n - 2j} < n!^{2n - 2j + 1}\] Thus, $ (a_i - a_j)|(n!^{2n - 2j + 2})|a_j$ as desired. This set works, so we are done.

We use induction. The base case is trivial. Let the elements of our set be $ a_1, a_2, \ldots, a_n$. Note that if we scale all the elements by some amount, the condition is still satisfied. Hence, we may scale all the variables by $ m = \prod_{1 \leq i \leq n} a_i \prod_{1 \leq j < k \leq n} (a_i - a_j)^2$. It follows that $ m$ and all numbers of the form $ a_i$ and $ a_j - a_k$ for all $ i$, $ j$, $ k$, $ j \neq k$, between 1 and $ n$ inclusive have the same set of prime factors. Let $ c = m + 1$. We claim that $ m$ and all numbers of the form $ c - a_i$, $ 1 \leq i \leq n$, are relatively prime. If $ p | m$ and $ p |c - a_i$ for some $ i$, then $ p | m + 1 - a_i$. Since $ m$ and $ a_i$ have the same set of prime factors, $ p | 1$, a contradiction. If $ p | c - a_i, c - a_j$ for some distinct $ i$ and $ j$, $ p | a_i - a_j$ as well. But $ a_i - a_j$ and $ a_i$ have the same prime factors, so $ p | a_i$. Therefore, $ p | c = m + 1$. But $ p | m$ as $ m$ and $ a_i$ have the same set of prime factors, so $ p | 1$, again a contradiction. Hence, by the Chinese remainder theorem, we can find some $ k$ such that $ k \equiv 0 \pmod m$, and $ k \equiv -a_i \pmod{(c - a_i)^2}$ for all integers $ i$ with $ 1 \leq i \leq n$. Let $ b_i = a_i + k$, and let $ b_{n+1} = c + k$. For all $ i, j$ with $ 1 \leq i < j \leq n$, $ (b_i - b_j)^2 = (a_i - a_j)^2$, and $ b_i b_j = (a_i + k)(a_j + k) = a_i a_j + k(a_i + a_j + k)$. Since $ (a_i - a_j)^2 | m | k$, and our inductive hypothesis tells us that $ (a_i - a_j)^2 | a_i a_j$, we have that $ (b_i - b_j)^2 | b_i b_j$. It remains to show that $ (b_{n+1} - b_i)^2 | b_{n+1} b_i$. But $ k$ was constructed so that $ (b_{n+1} - b_i)^2 = (c - a_i)^2 | a_i + k = b_i | b_i b_{n+1}$, so our proof is complete.

We induct on $n$. For the base case $n=2,$ just pick two numbers which differ by 1. Let $a_1, a_2, \cdots , a_j$ be $j$ numbers which satisfy $(a_m-a_n)^2 \mid a_ma_n \quad \forall \ m \neq n.$ Let $M = \displaystyle \prod_{m \neq n} (a_m - a_n)^2 \cdot a_1^2 a_2^2 \cdots a_j^2.$ Notice that the set $\{M+a_1, M+a_2, \ldots, M+a_j\}$ satisfies the property too since for any $m \neq n,$ \[(a_m-a_n)^2= ((M+a_m) - (M+a_n))^2 \mid a_ma_n \\ \implies ((M+a_m) - (M+a_n))^2 \mid (M+a_m)(M+a_n).\] We need to append another number to this set. Appending $M$ to this set works, since $((M+a_i)-M)^2 = a_i^2 \mid M(M+a_i) \quad \forall \ i.$ Hence, we have found a $j+1$ element set of integers with the required property. By induction, we are done. $\blacksquare$