Maximum number dividing $\prod_{iFix an integer $n$. What is the maximum number guaranteed to divide $\prod_{i<j}(a_i-a_j)$ for any integers $a_1,\ldots,a_n$?
For instance, if $n=3$, then two of the three numbers have the same parity, hence the product is divisible by $2$. We have $(0-1)(0-2)(1-2)=-2$, so the answer is $2$.
For $n=4$, two of the four numbers are the same modulo $3$. Also, either two are odd and two are even, or three have the same parity, so the product is divisible by $4$. We have $(0-1)(0-2)(0-3)(1-2)(1-3)(2-3)=12$, so the answer is $12$.
 A: Partial solution The number you seek is a multiple of $lcm[1,2,3,..., n-1]$, where $lcm$ denotes the least common multiple.
Proof: $lcm[1,2,3,..., n-1]$ always divides $\prod_{i<j}(a_i-a_j)$:
By the pigeon hole principle, for each $1 \leq k \leq n-1$ you can find $i \neq j$ so that $k | (a_i-a_j)$. Therefore $k| \prod_{i<j}(a_i-a_j)$.
This shows that $\prod_{i<j}(a_i-a_j)$ is a multiple of $1,2,3,.., n-1$, and therefore a multiple of their lcm.

Now let $N > lcm[1,2,3,..., n-1]$ be the number  you are seeking. Since $N$ and $lcm [1,2,3,...,n-1]$ both divide $\prod (a_i-a_j)$ so does $lcm(N, lcm[1,2,3,.., n-1])$
Since $N$ is maximal with this property, we get
$$lcm(N, lcm[1,2,3,.., n-1])=N$$
which proves the above claim.

Next, it is easy to prove that all primes appearing in the prime factorization of $N$ must be less or equal than $n-1$. But, many of these primes have higher powers.
The problem now is that for small primes $p$, you will get too many pairs $a_i, a_j$ so that $p|(a_j-a_i)$ besides the ones which are divisible by the last $p^k \leq n-1$.
For example, when $n=6$ at least six pairs  of numbers must have the same parity(when there 3 odd, 3 even numbers, for any other choice we have more pairs), and by pigeonhole principle, at least a pair must have the difference divisible by $4$. This means than when $n=6$ the power of $2$ in $N$ is actually $7$.
Last but not least, if $\frac{n}{2} <p < n$ it is easy to prove that the power of $p$ in $N$ is exactly 1.
In general, for each given $n$ you know all the primes appearing in the prime factorization, and you can calculate the power of each of the primes in $N$ as I calculated for $2$ in $6$, looking for the worst case scenario. This seems to be a problem for the generalized pigeon hole principle:
if $kp^s+1 \leq n-1 < (k+1)p^s+1$, I think you get at least $(k+1)p$ pairs each with the difference divisible by $p^s$. Repeat for all powers $p^s$ less than $n-1$.
