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
$$
17\# = 2\cdot 3\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17 = 510\,510
$$
The correct value should be $y = 770$.
You have almost solved the problem. Once you see that the sum is equal to $(2^2 \times 5 \times 7 \times 11)/xy,$ you can just set $x = 2$ so that $y = 11 \times 7 \times 5 \times 2 = 770$. This satisfies the desired constraints, and it's clear that we can't do any better (we've exhausted all of the prime factors available, and changing the value of $x$ will only decrease the value of $y$).
Note that the expression is still an integer:
$$\frac{2^2 \times 5 \times 7 \times 11}{xy} = \frac{1540}{2 \times 770} = 1.$$
Best Answer
First every such integer is divisible by $7, 11, 13$.
Highest power of $2$ of $13!$ is even, so you need to find an integer product such that it's odd. This can be easily done.
Then the highest power of $3$ of $13!$ is odd, so you need to find an integer product such that it's divisible by $3$ but not by $9$. This is easy too.
The highest power of $5$ of $13!$ is $2$. It's easy to find one integer product that's not divisible by $5$.
Therefore the $\gcd$ of all integer products is $3\times 7 \times 11 \times 13 = 3003$.