In the Cartesian coordinate system $x o y$, the coordinates of point $A\left(x_{1}, y_{1}\right)$ and point $B\left(x_{2}\right.$, $y_{2}$) are both one-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 $\triangle O A^{\prime} A$, form a four-digit number from $x_{1}, y_{1}, x_{2}, y_{2}$, and write out the solution process.
Given $S_{\triangle B O B^{\prime}}=S_{\triangle A O A^{\prime}}+33.5$,
we have
$$
\begin{array}{c}
x_{2} y_{2}=x_{1} y_{1}+6.7 . \\
x_{1} y_{1}>0, \quad \therefore x_{2} y_{2}>67 .
\end{array}
$$
Since $x_{2}, y_{2}$ are positive single-digit integers,
$$
\therefore x_{2} y_{2}=72 \text {, or } 81 .
$$
But $\because \angle B O B^{\prime}y_{2} \text {, }$
$$
\therefore x_{2} y_{2} \neq 81 \text {, }
$$
Thus,
$$
x_{2} y_{2}=72 \text {. }
$$
$$
x_{2}=9, y_{2}=8 \text {. }
$$

Therefore,
$$
\text { Also, } \because \quad \angle A O B>45^{\circ}, \quad \therefore \quad x_{1}<y_{1} \text {, }
$$

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