Three, (This question is worth 16 points) How many ordered pairs of positive integers $(x, y)$ have the following properties:
$y<x \leqslant 100$, and $\frac{x}{y}$ and $\frac{x+1}{y+1}$ are both integers?
Three, let $\frac{x+1}{y+1}=m(m \in \mathbf{N}, m>1)$, then $x=m y+(m-1)$.
Since $y \mid x$, it follows that $y \mid(m-1)$.
Let $m-1=k y(k \in \mathbf{N})$, then $m=k y+1$,
Substituting into (1), we get
$$
\begin{array}{l}
x=(k y+1) y+k y=k y(y+1)+y \leqslant 100, \\
k \leqslant \frac{100-y}{y(y+1)} .
\end{array}
$$

Thus, there are $\left[\frac{100-y}{y(y+1)}\right]$ positive integers $k$ that satisfy the above inequality.

Conversely, for each positive integer $y$, equation (2) shows a one-to-one correspondence between $k$ and the ordered pair $(x, y)$, and makes
$$
\frac{x}{y}=k(y+1)+1, \frac{x+1}{y+1}=k y+1 .
$$

Notice that, when $10 \leqslant y \leqslant 99$,
$$
y(y+1) \geqslant 110>100-y,\left[\frac{100-y}{y(y+1)}\right]=0,
$$

The number of ordered pairs is
$$
\begin{array}{l}
\sum_{y=1}^{99}\left[\frac{100-y}{y(y+1)}\right]=\sum_{y=1}^{9}\left[\frac{100-y}{y(y+1)}\right] \\
=49+16+8+4+3+2+1+1+1=85 .
\end{array}
$$