In the Cartesian coordinate system $x O y$, the coordinates of points $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$ are both single-digit positive integers, the angle between $O A$ and the positive direction of the $x$-axis is greater than $45^{\circ}$, the angle between $O B$ and the positive direction of the $x$-axis is less than $45^{\circ}$, the projection of $B$ on the $x$-axis is $B^{\prime}$, the projection of $A$ on the $y$-axis is $A^{\prime}$, the area of $\triangle O B^{\prime} B$ is 33.5 larger than the area of $\triangle O A^{\prime} A$, and the four-digit number formed by $x_{1}, y_{1}, x_{2}, y_{2}$ is 'S'.
1985
I. 【Analysis and Solution】From the known $S_{\triangle B O B^{\prime}}=S_{\triangle A O A^{\prime}}+33.5$, we get $x_{2} y_{2}=x_{1} y_{1}+67 . x_{1} y_{1}>0, \therefore x_{2} y_{2}>67$.
$\because x_{2} y_{2}$ is a positive integer.
$$
\therefore x_{2} y_{2}=72 \text { or } 81 \text {. }
$$

But $\because \angle B O B^{\prime}y_{2}
$$

Therefore, $x_{2} y_{2} \neq 81$,
Thus, $x_{2} y_{2}=72 . \therefore x_{2}=9, y_{2}=8$.
Hence, $x_{1} y_{1}=5$.
$$
\text { Also, } \because \angle A O A^{\prime}>45^{\circ}, \therefore x_{1}<y_{1} \text {, }
$$

Since both are single-digit integers, we deduce $x_{1}=1, y_{1}=5$.
$$
\therefore \overline{x_{1} x_{2} y_{1} y_{2}}=1985 .
$$