Let $s$ be the sidelength of the original square, $m$ and $n$ the numbers of white and green rectangles $W_{1}, W_{2}, \ldots, W_{m}$ and $G_{1}, G_{2}, \ldots, G_{n}$, respectively, $a_{i}$ and $b_{i}$ the width and height of the white rectangle $W_{i}$, respectively, for $i=1,2, \ldots, m$, and $c_{j}$ and $d_{j}$ the width and height of the green rectangle $G_{j}$, respectively, for $j=1,2, \ldots, n$.

The total area of the white rectangles and the total area of the green rectangles satisfy
$$a_{1} b_{1}+a_{2} b_{2}+\cdots+a_{m} b_{m}=c_{1} d_{1}+c_{2} d_{2}+\cdots+c_{n} d_{n}=\frac{s^{2}}{2}$$
We have
$$\begin{aligned} S & =\frac{a_{1}}{b_{1}}+\frac{a_{2}}{b_{2}}+\cdots+\frac{a_{m}}{b_{m}}+\frac{d_{1}}{c_{1}}+\frac{d_{2}}{c_{2}}+\cdots+\frac{d_{n}}{c_{n}} \\ & =\frac{a_{1}^{2}}{a_{1} b_{1}}+\frac{a_{2}^{2}}{a_{2} b_{2}}+\cdots+\frac{a_{m}^{2}}{a_{m} b_{m}}+\frac{d_{1}^{2}}{c_{1} d_{1}}+\frac{d_{2}^{2}}{c_{2} d_{2}}+\cdots+\frac{d_{n}^{2}}{c_{n} d_{n}} \\ & \geq \frac{\left(a_{1}+a_{2}+\cdots+a_{m}\right)^{2}}{\frac{s^{2}}{2}}+\frac{\left(d_{1}+d_{2}+\cdots+d_{n}\right)^{2}}{\frac{s^{2}}{2}} \end{aligned}$$
by Cauchy-Schwarz inequality.

Assume that $a_{1}+a_{2}+\cdots+a_{m}<\frac{s}{2}$. Because $b_{i} \leq s$, for $i=1,2, \ldots, m$, we have
$$\frac{s^{2}}{2}=a_{1} b_{1}+a_{2} b_{2}+\cdots+a_{m} b_{m} \leq\left(a_{1}+a_{2}+\cdots+a_{m}\right) s<\frac{s^{2}}{2},$$
which is a contradiction. Therefore $a_{1}+a_{2}+\cdots+a_{m} \geq \frac{s}{2}$ and, similarly, $d_{1}+d_{2}+\cdots+d_{n} \geq \frac{s}{2}$.

Now assume that $a_{1}+a_{2}+\cdots+a_{m}<s$. There exists a vertical line $L$ crossing the original square and not intersecting any of the white rectangles. Therefore, the sum of the heights of the green rectangles intersecting $L$ is greater than or equal to $s$. Hence, either $a_{1}+a_{2}+\cdots+a_{m} \geq s$ or $d_{1}+d_{2}+\cdots+d_{n} \geq s$.

We deduce that
$$S \geq \frac{2}{s^{2}}\left(s^{2}+\left(\frac{s}{2}\right)^{2}\right)=\frac{5}{2}$$
If Turki divides the square into two horizontal rectangles of the same area then $S=\frac{2}{1}+\frac{1}{2}=\frac{5}{2}$. Thus the minimum possible value of $S$ is $\frac{5}{2}$.