One, (20 points) A research study group from Class 1, Grade 3 of Yuhong Middle School conducted a survey on students' lunch time at the school canteen. It was found that within a unit of time, the number of people buying lunch at each window and the number of people choosing to eat outside due to unwillingness to wait for a long time are both fixed numbers. Furthermore, it was found that if only 1 window is opened, it takes $45 \mathrm{~min}$ to ensure that all waiting students can buy lunch; if 2 windows are opened simultaneously, it takes $30 \mathrm{~min}$. It was also discovered that if all waiting students can buy lunch within $25 \mathrm{~min}$, the number of people eating outside can be reduced by $80 \%$ within a unit of time. Assuming the total number of students in the school remains constant and everyone needs to have lunch, to facilitate student dining, the research group suggests that the school canteen should sell out lunch within $20 \mathrm{~min}$. How many windows should be opened simultaneously at least?
Let each window sell lunch to $x$ people per minute, and $y$ people go out to eat per minute, with the total number of students being $z$ people, and assume that at least $n$ windows need to be open simultaneously. According to the problem, we have
$$
\left\{\begin{array}{l}
45 x=z-45 y, \\
2 \times 30 x=z-30 y, \\
20 n x \geqslant z-20(1-80 \%) y .
\end{array}\right.
$$

From (1) and (2), we get $y=x, z=90 x$. Substituting into (3) yields
$$
20 n x \geqslant 90 x-4 x, n \geqslant 4.3 \text{. }
$$

Therefore, at least 5 windows need to be open simultaneously.