Three. (25 points) Let the pairs of positive integers $(m, n)$, where both are no more than 1000, satisfy
$$
\frac{m}{n+1}<\sqrt{2}<\frac{m+1}{n} \text {. }
$$

Find the number of all such pairs $(m, n)$.
Three, 1706.
$$
\begin{array}{l}
\text { Given } \frac{m}{n+1}<\sqrt{2}<\frac{m+1}{n} \\
\Rightarrow \sqrt{2} n-1<m<\sqrt{2}(n+1) .
\end{array}
$$

For each $n$, the number of integers in the above range is
$$
\begin{array}{l}
{[\sqrt{2}(n+1)]-[\sqrt{2} n-1]} \\
=[\sqrt{2}(n+1)]-[\sqrt{2} n]+1,
\end{array}
$$

where $[x]$ denotes the greatest integer not exceeding the real number $x$.
And $707 \sqrt{2}<1000<708 \sqrt{2}$
$$
\Rightarrow n \leqslant 707 \text {. }
$$

When $n=707$, $m=999,1000,1001$, but 1001 should be discarded.
Therefore, the number of pairs $(m, n)$ that satisfy the condition is
$$
\begin{array}{l}
\sum_{n=1}^{706}([\sqrt{2}(n+1)]-[\sqrt{2} n]+1)+2 \\
=\sum_{n=1}^{706}([\sqrt{2}(n+1)]-[\sqrt{2} n])+708 \\
=708+[707 \sqrt{2}]-[\sqrt{2}] \\
=708+999-1=1706 .
\end{array}
$$
(Chen Qian, Chen Hongfei, Yuyan Middle School, Xishui County, Huanggang City, Hubei Province, 438200)