8.8. In a $10 \times 10$ grid, natural numbers 1 to 100 are written: the first row from left to right contains $1 \sim 10$; the second row from left to right contains $11 \sim 20$; and so on. Andrei tries to divide the entire grid into $1 \times 2$ rectangles, calculate the product of the two numbers in each rectangle, and then add up all the products. How should he divide the grid to make the sum as small as possible?
8.8 . A rectangle of size $1 \times 2$ is called a "domino". Number the dominoes. Let the two numbers written in the $i$-th domino be $a_{i}$ and $b_{i}$, then
$$
a_{i} b_{i}=\frac{a_{i}^{2}+b_{i}^{2}}{2}-\frac{\left(a_{i}-b_{i}\right)^{2}}{2} .
$$

Write such an expression for each domino. After summing, it is found that the sum $S$ of the 50 products to be examined is
$$
\begin{aligned}
S= & \frac{a_{1}^{2}+a_{2}^{2}+\cdots+a_{50}^{2}+b_{1}^{2}+b_{2}^{2}+\cdots+b_{50}^{2}}{2}- \\
& \frac{\left(a_{1}-b_{1}\right)^{2}+\left(a_{2}-b_{2}\right)^{2}+\cdots+\left(a_{50}-b_{50}\right)^{2}}{2} .
\end{aligned}
$$

The first fraction on the right side of the above equation equals $\frac{1^{2}+2^{2}+\cdots+100^{2}}{2}$, whose value is independent of the partitioning method. The terms in the numerator of the second fraction are either $1^{2}$ or $10^{2}$, depending on whether the domino is placed horizontally or vertically.

Therefore, when all dominoes are placed vertically, each term is 100, the second fraction reaches its maximum, and at this point, $S$ is the smallest.