You can rewrite this as: $\frac{m}{\sqrt{2}}-1
Since $\sqrt{2}$ is irrational, we know that $\frac{m}{\sqrt{2}}-1$ is not an integer, so for an integer $n>\frac{m}{\sqrt{2}}-1$ iff $n\geq \left\lceil \frac{m}{\sqrt{2}}-1\right\rceil =\left\lfloor \frac{m}{\sqrt{2}}\right\rfloor$.
Similarly, $n<\frac{m+1}{\sqrt{2}}$ iff $n\leq \left\lfloor \frac{m+1}{\sqrt{2}}\right\rfloor$. So altogether, we are seeking $m,n$ such that $\left\lfloor\frac{m}{\sqrt 2}\right\rfloor\leq n\leq \left\lfloor\frac{m+1}{\sqrt{2}}\right\rfloor$
For a particular $m$, then, the number of possible $n$ is: $\left\lfloor\frac{m+1}{\sqrt{2}}\right\rfloor-\left\lfloor\frac{m}{\sqrt 2}\right\rfloor +1$. Summing over all $m$, the result is $-1+\sum_{m=1}^{1000}\left(\left\lfloor\frac{m+1}{\sqrt{2}}\right\rfloor-\left\lfloor\frac{m}{\sqrt 2}\right\rfloor +1\right)$
(The $-1$ is because we don't want to count $n=0$ when $m=1$.)
But this is just $999$ plus a telescoping sum, and we see that the result is:
$999+\left\lfloor\frac{1001}{\sqrt{2}}\right\rfloor$
Actually, even more specifically, it is:
$1000-\lfloor\sqrt{2}\rfloor + \left\lfloor\frac{1000+1}{\sqrt{2}}\right\rfloor$
This will work for any irrational number $\alpha>1$ and any upper bound, $M>\alpha$, yielding a total: $M-\lfloor\alpha\rfloor + \left\lfloor\frac{M+1}{\alpha}\right\rfloor$
which counts the pairs $(m,n)$ with $1\leq m,n\leq M$ and $\frac{m}{n+1}<\alpha<\frac{m+1}n$