How to sum $ \displaystyle\sum_{n=1}^\infty \sum_{\substack{j=1 \\\mathrm{gcd}(2n,j) = 1}}^{\lfloor n\sqrt{2}\rfloor} f(n,j) $

As the title suggests, I am trying to work out a double sum where one of the indexes has a condition of coprimality, meaning the index doesn't sum whenever $\mathrm{gcd}(2n,j) \ne 1$. The sum is the following:

$$\sum_{n=1}^\infty \sum_{\substack{j=1 \\\mathrm{gcd}(2n,j) = 1}}^{\lfloor n\sqrt{2}\rfloor} \left( \frac{1}{4n^2(n+j)^2} - \frac{1}{j^2(2n+j)^2} \right)$$

I already know more or less how to deal with these kind of sums when $j$ goes from $1$ to $\infty$, but when the sum is finite I have absolutely no idea (and the fact that the index goes up to $\lfloor n\sqrt{2}\rfloor$ doesn't help either).

Numerical evidence suggests this sum is approximately $-0.16639$.