8. (10 points) Divide the six natural numbers $1 \sim 6$ into two groups, Jia and Yi. The maximum product of the sum of the numbers in group Jia and the sum of the numbers in group Yi is . $\qquad$
【Analysis】To maximize the product of the sum of numbers in group A and the sum of numbers in group B, these two sums should be as large as possible, and the difference between the two sums should be as small as possible; thus, the groups should be divided accordingly.

【Solution】Solution: The sum of the 6 numbers is:
$$
1+2+3+4+5+6=21 \text {, }
$$
21 can be decomposed into the sum of two closest integers as: $21=11+10$,
that is, the six natural numbers from $1 \sim 6$ can be divided into two groups: $(1+2+3+4)$ and $(5+6)$,
Therefore, the maximum product is:
$$
\begin{aligned}
& (1+2+3+4) \times(5+6) \\
= & 11 \times 10 \\
= & 110 ;
\end{aligned}
$$

Answer: The maximum product of the sum of numbers in group A and the sum of numbers in group B is 110.
Unit: 110.