For $n \in \mathbb{N}$ and $n>(q+1)/4$. Where $q>1$ and integer. With $p\in \mathbb{Z}$
$\frac{(4n-1)!}{q}p - \frac{(4n-1)!}{3} -4(4n-1)! \big(\frac{1}{4n-3}- \frac{1}{4n-1}+\frac{1}{1}- \frac{1}{3}+\frac{1}{5}-\frac{1}{7}+... \big)$
I suspect that it's not integer. $1-1/3+1/5-...$ converges to $\pi/4$. And the rest of the sums are integers.
Let $n=1, q=2, p=3$. Then $4n>q+1$ but $$\frac{(4n-1)!}{q}p - \frac{(4n-1)!}{3} -4(4n-1)! \left(\frac{1}{4n-3}- \frac{1}{4n-1}+\frac{1}{1}- \frac{1}{3}+\frac{1}{5}-\frac{1}{7}+... \right)$$ becomes $$9-2-24\left(1-\frac13+\frac\pi4\right)=7-24+8-6\pi=-9-6\pi\not\in\mathbb{Z}$$ so the result does not hold.