Nate's hint (that $n^4 \equiv 0,1 \mod 16$ depending on whether $n$ is even or odd, respectively) gives you the answer. Since $65536\equiv 0 \mod 16$, if any of the $n_i$ are odd, then all of the $n_i$ must be odd so that their sum is $16\equiv 0 \mod 16$. But there are no two consecutive odd numbers. If you then look at all $n_i$ being even, you have the same problem. So there is no solution containing two consecutive values of $n_i$.
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
$$
17\# = 2\cdot 3\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17 = 510\,510
$$
Best Answer
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.