4. Among the positive integers not greater than 100, the ordered integer pairs $(m, n)$ that satisfy
$$
\frac{m}{n+1}<\sqrt{2}<\frac{m+1}{n}
$$

are $\qquad$ pairs.
4. 170 .

Notice that $\sqrt{2} n-1<m<\sqrt{2}(n+1)$.
For each $n$, the number of $m$ is given by
$$
\begin{array}{l}
{[\sqrt{2}(n+1)]-[\sqrt{2} n-1]} \\
=[\sqrt{2}(n+1)]-[\sqrt{2} n]+1
\end{array}
$$

Since $100<71 \sqrt{2}<101<72 \sqrt{2}$, we have $n \leqslant 71$.
But when $n=71$, $m=100$.
Therefore, the number of $m$ is
$$
\begin{array}{l}
\sum_{n=1}^{70}([\sqrt{2}(n+1)]-[\sqrt{2} n]+1)+1 \\
=[71 \sqrt{2}]-[\sqrt{2}]+71 \\
=100-1+71=170 .
\end{array}
$$