The hint.
Prove that $$\frac{a}{b}+\frac{c}{d}=\frac{d}{c}+\frac{c}{d}\leq-2$$ or
$$\frac{b}{c}+\frac{d}{a}=\frac{a}{d}+\frac{d}{a}\leq-2$$ and use
$$\frac{a}{c}+\frac{b}{d}+\frac{c}{a}+\frac{d}{b}=\left(\frac{a}{b}+\frac{c}{d}\right)\left(\frac{b}{c}+\frac{d}{a}\right).$$
A full solution.
Let $\frac{a}{b}+\frac{c}{d}>0$ and $\frac{b}{c}+\frac{d}{a}>0$.
Thus, by AM-GM $$4=\left(\frac{a}{b}+\frac{c}{d}\right)+\left(\frac{b}{c}+\frac{d}{a}\right)=\left(\frac{d}{c}+\frac{c}{d}\right)+\left(\frac{a}{d}+\frac{d}{a}\right)\geq2+2=4,$$
which gives $c=d$ and $a=d$, which is impossible.
Thus, one of the expressions $\frac{a}{b}+\frac{c}{d}$ or $\frac{b}{c}+\frac{d}{a}$ is negative.
Let $u=\frac{a}{b}+\frac{c}{d}<0$.
Thus, $$u=\frac{a}{b}+\frac{c}{d}=\frac{d}{c}+\frac{c}{d}\leq-2$$ and we need to prove that
$$u(4-u)\leq-12$$ or
$$(u+2)(u-6)\geq0,$$ which is obvious.
The equality occurs for $\frac{a}{b}+\frac{c}{d}=-2$, $\frac{b}{c}+\frac{d}{a}=6$ and $ac=bd.$
Easy to see that it's possible, which says that $-12$ is a maximal value.
$4$ squares on a $n \times n$ would create a parallelogram with their centers if the squares are at coordinates of the form $(i_1, j_1), (i_1, j_1+k), (i_2, j_2), (i_2, j_2+k)$ ($i_1\neq i_2$) or of the form $(i_1,j_1),(i_1+k,j_1),(i_2,j_2),(i_2+k,j_2)$ ($j_1\neq j_2$). [it's not a necessary condition - only a sufficient one]
Denote by $m$ the minimal number of squares such that every subset of $m$ squares contain a parallelogram.
It is easy to see that $m \ge 2n$ because there isn't a parallelogram among $(1,1),...,(1,n), (2,1), (3,2),..., (n,n-1)$.
I will prove that $m=2n$:
Consider a subset of $2n$ squares from the $n^2$ squares, and let $a_1,...,a_n$ denote the number of squares in rows $1,...,n$ respectively.
Notice that in the $i$'th row, there are at least $a_i-1$ unique distances between chosen squares in that row (the leftmost with each of the others). If we would repeat the same distance in $2$ distinct rows, it would create a parallelogram, so all of those distances are unique.
So we got $\sum_{i=1}^{n} {a_i-1} = n$ unique distances, but each distance is in $\{1,...,n-1\}$ - a contradiction.
Best Answer
We want the $\gcd$ of all the different products $$ 1\cdot (1\cdot 2\cdot 3\cdot 4\cdots 16\cdot 17)\\ 2\cdot (1\cdot 3\cdot 5\cdot 7\cdots 31\cdot 33)\\ 3\cdot (1\cdot 4\cdot 7\cdot 10\cdots 46\cdot 49)\\ \vdots\\ 2016\cdot (1\cdot 2017\cdot 4033\cdot 6049\cdots 30\,241\cdot 32\,257) $$ (I added a factor $1$ to make sure that we had $17$ factors in each parenthesis. This makes $17$ not a special case, which makes the proof easier.)
For each prime $p\leq 17$, consider $P(p)$. It has $p$ as the first factor, and then seventeen factors that are coprime to $p$. So $p\mid P(p)$, but $p^2\nmid P(p)$. Similarily, for any $n$ with $p\mid n$, we have $p\mid P(n)$ as $n$ is a factor of $P(n)$.
For any other $n$, $P(n)$ has $p$ as a factor because the part of the product above that I have put into brackets is an arithmetic progression of length $17\geq p$, with common difference coprime to $p$, and at least one of the terms must therefore be divisible by $p$. So the $\gcd$ contains $p$ as a factor, but not $p^2$.
As for any prime factors above $17$, none of them appear in $P(1)$, so they do not divide the $\gcd$.
We conclude that the $\gcd$ of all the above products is $$ 17\# = 2\cdot 3\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17 = 510\,510 $$