Lemma 1 If $ \gcd(a,b) = m$ and $ a = mk, b = ml$, then $ (a-b)^{2}|ab$ iff $ k-l = 1$. Proof $ (a-b)^{2}= m^{2}(k-l)^{2}$; $ ab = m^{2}kl$ so $ (a-b)^{2}|ab\Leftrightarrow (k-l)^{2}|kl$. But because$ \gcd(a,b) = m$and$ a = mk, b = ml$,$ \gcd(k,l) = 1$and if $ k-l = q > 1$ then $ k-l$ is divisible by a prime$ p\geq 2$, and then$ p$divides$ k$or$ l$, wolog$ p|k$, but then as $ p|(k-l)$ then $ p|l$ which contradicts the fact that $ \gcd(k,l) = 1$. So,$ k-l = 1$and the lemma is proved in one direction. Now, $ (a-b)^{2}= (mk-ml)^{2}= m^{2}(k-l)^{2}= m^{2}|m^{2}kl = (mk)(ml) = ab$ so $ (a-b)^{2}|ab$ and the lemma is proved in the other direction. Lemma 2 If $ x,y$ are two distinct positive integers with $ x > y$ such that $ (x-y)^{2}|xy$ then for any positive integer $ c\geq 2$, the numbers $ t = c^{x}-1, s = c^{x-y}(c^{y}-1)$ then $ (s-t)^{2}|st$. Proof By lemma 1, if $ \gcd(x,y) = m$ and $ x = mk,y = ml$ then $ k-l = 1$. Then, $ t = c^{x}-1 = c^{mk}-1 = (c^{m}-1)(c^{(k-1)m}+c^{(k-2)m}+...+c^{m}+1)$ Also, $ s = c^{x-y}(c^{y}-1) = c^{mk-ml}(c^{ml}-1) = c^{m(k-l)}(c^{m}-1)(c^{(l-1)m}+c^{(l-2)m}+...+c^{m}+1)$ $ = (c^{m}-1)(c^{m}(c^{(l-1)m}+c^{(l-2)m}+...+c^{m}+1)) = (c^{m}-1)((c^{lm}+c^{(l-1)m}+...+c^{2m}+c^{m}))$ $ = (c^{m}-1)(c^{(k-1)m}+c^{(k-2)m}+...+c^{m})$ because $ k-l = 1$. Therefore, because we can find numbers $ d = c^{m}-1, e = c^{(k-1)m}+c^{(k-2)m}+...+c^{m}+1, f = c^{(k-1)m}+c^{(k-2)m}+...+c^{m}$ such that $ t = de, s = df, e-f = 1$ then by lemma 1 $ (s-t)^{2}|st$ and lemma 2 is proved. Finally note that for the numbers $ c^{z}= c^{z-u}(c^{u})$ and $ c^{z-u}(c^{u}-1)$ where $ c$ is a positive integer greater than $ 1$ there exist numbers $ d = c^{z-u}, e = c^{u}, f = c^{u}-1$ such that $ c^{z}= de, c^{z-u}(c^{u}-1) = df, e-f = 1$ then by lemma 1 $ (c^{z}-c^{z-u}(c^{u}-1))^{2}|(c^{z}*c^{z-u}(c^{u}-1))$ -- call this property lemma 3. Now, we can construct the set $ S$ inductively. Start with numbers $ 1,2$ as the base step for $ n = 2$. Then, when we have $ k$ numbers $ a_{1}< a_{2}< ... < a_{k}$ belonging to set $ S$ and satisfying the conditions, we can construct a new set of $ k+1$ numbers: $ b_{1}< b_{2}< ...b_{k+1}$ where for $ i = 1,2,...,k$, $ b_{i}= r^{a_{k}-a_{i}}(r^{a_{i}}-1)$ and $ b_{k+1}= a^{k}$ where $ r$ is any positive integer greater than $ 1$. Then, for any $ i,j\in\{1,2,...,k\}$, $ i < j$ by lemma 1, $ (b_{j}-b_{i})^{2}|b_{i}b_{j}$, and for any $ i\in\{1,2,...,k\}$, by lemma 3, $ (b_{k+1}-b_{i})^{2}|b_{k+1}b_{i}$ so the constructed set of $ k+1$ numbers satisfies the conditions of the problem. Then, from the new set with $ k+1$ numbers we can construct a set with $ k+2$ numbers and so on, until we reach $ n$, and the result follows.

I have am other method . Lemma 1 $ A = \{a\}$ and $ |A| = k$ satisfy $ (a - b)^2|ab,\forall a,b\in A$ then the set $ A_1 = \{a + d\}$ has this property where $ \prod_{i > j = 1}^k (a_i - a_j)^2|d$ satisfy. Lemma 2 If the set $ A$ satisfy the condition then the set $ A + \{0\}$ also satisfy the condition. It is quite easy to prove . Now we prove by induction: $ k = 2$ then chose $ a,b = 1,2$ Suppose it is true for $ k$ we prove it is also true for $ k + 1$ From lemma 1,lemma 2 then it true . With this method we can solve problem. Let $ a,b\in N$ satisfy condition $ (a - b)^2|ab$ then exist A a set such that 1)$ a,b\in A$ 2)$ (x - y)^2|xy,\forall x,y\in A$

Assume the set is finite. Let B be the lcm of every element in the set. Add 0 to the set. Increase each element by B. We now have a larger set so there is no maximum and the set can be infinite. edit: So we can make a set of any finite size. Luckily this is all the problem was asking for and we dont need it to be an infinite set.

We must be careful! As stated above, "We now have a larger set so there is no maximum and the set can be infinite.", this was yet unproven, and, moreover, false, as I will show below. All that was proven is that from a set with $n$ elements we can create one with $n+1$ elements, and so we can create as large a set as wanted, but not necessarily an infinite set. Consider any such set $S$. Take an element $a \in S$, $a \neq 0$. Then for all elements $b\in S$, $b \neq a$, we will need have $(a-b)^2 \mid ab$, hence $(a-b)^2 \leq |ab|$, therefore $b^2 \leq |ab| + 2ab - a^2 < 3|ab|$. So, $|b| < 3|a|$, therefore $|S| \leq 6|a|$.