13. (15 points) As shown in the figure, ten points are evenly marked on the circumference of a circle. Place the ten natural numbers $1 \sim 10$ on these ten points. Rotate a line passing through the center of the circle. When the line does not pass through any marked points, it divides the numbers 1 to 10 into two groups. For each arrangement, there are five ways to divide the numbers as the line rotates. For each division, there is a product of the sums of the two groups, and the smallest of these five products is denoted as $K$. What is the maximum value of $K$ among all possible arrangements?
【Analysis】No matter how these 10 numbers are placed, their sum remains unchanged, $1+2+3+4+5+6+7+8+9+10=55$, divided into two groups, the closer the sums of the two groups, the larger the required product will be.

【Solution】
$$
\begin{array}{l}
1+2+3+4+5+6+7+8+9+10=55 \\
55=27+28
\end{array}
$$

Therefore, as long as the 10 numbers are divided into two parts with sums of 27 and 28, the product at this time is $27 \times 28=756$. Answer: In all the arrangements, $K$ is maximized at 756.