\%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 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 more than the area of $\triangle O A^{\prime} A$. The four-digit number $\overline{x_{1} x_{2} y_{2} y_{1}}=x_{1} \cdot 10^{3}+x_{2} \cdot 10^{2}+y_{2} \cdot 10 + y_{1}$. Try to find all such four-digit numbers, and write out the solution process.
As shown in the figure, from the given conditions, we know that $x_{2} y_{2} - x_{1} y_{1} = 67$. Since $x_{1} > 0, y_{1} > 0$, it follows that $x_{2} y_{2} > 67$. Also, because $x_{2}$ and $y_{2}$ are both one-digit positive integers, we have: $x_{2} y_{2} = 8 \times 9 = 72$ or $x_{2} y_{2} = 9 \times 9 = 81$. From $\angle B O B^{\prime} y_{2}$, we know that $x_{2} y_{2} \neq 81$, thus $x_{2} y_{2} = 9 \times 8 = 72$, which means $x_{2} = 9, y_{2} = 8$. Therefore, $x_{1} y_{1} = x_{2} y_{2} - 67 = 5$. Since $\angle A O B^{\prime} > 45^{\circ}$, it follows that $x_{1} < y_{1}$. Given that $x_{1}$ and $y_{1}$ are one-digit positive integers, the only possibility is $x_{1} = 1, y_{1} = 5$. Therefore, $\overline{x_{1} x_{2} y_{2} y_{1}} = 1985$.