9. There are 5 circular flower beds, with diameters of 3 meters, 4 meters, 5 meters, 8 meters, and 9 meters; please divide these 5 flower beds into two groups, to be managed by two classes, so that the areas managed by the two classes are as close as possible.
9.【Solution】We know that the area of each circle is equal to the square of the diameter multiplied by $(\pi / 4)$. Now we need to divide 5 circles into two groups, so that the total area of the two groups is as close as possible, or in other words, the ratio of the total areas of the two groups is as close to 1 as possible. Since each circle's area has a common factor of $(\pi / 4)$, and we are concerned with the ratio of the areas, we can eliminate this common factor, simplifying the problem to: divide the 5 circles into two groups so that the sum of the squares of the diameters of the two groups is as close as possible.
The squares of the diameters of the 5 circles are $9, 16, 25, 64, 81$.

The sum of these 5 numbers is 195. Since 195 is an odd number, it is impossible to divide these 5 numbers into two groups with equal sums. On the other hand, $81 + 16 = 97$, and $9 + 25 + 64 = 98$, which differ by only 1.

Therefore, the flower beds with diameters of 4 meters and 9 meters should be assigned to one class, and the other three flower beds should be assigned to another class.

Answer: The flower beds with diameters of 4 meters and 9 meters should be assigned to one class, and the other three flower beds should be assigned to another class.