Solve: To calculate the sum $S$ of these 27 sums, since each number appears in exactly 3 rows, we have
$$S=3 \times(1+2+\cdots+27)=3 \times 27 \times 14$$

Thus, $S$ is an even number, so the number of odd numbers among these 27 sums must be even.
If 26 of these 27 sums are odd, let's assume that the even number is the sum of the 3 numbers in the first row of Figure 1, i.e., $a_{1}+a_{2}+a_{3}$ is even, while the sums of the other 5 rows in Figure 1 are all odd. In this case, summing the numbers in Figure 1 by rows and columns, we get
$$\begin{aligned} & \left(a_{1}+a_{2}+a_{3}\right)+\left(a_{4}+a_{5}+a_{6}\right)+\left(a_{7}+a_{8}+a_{9}\right) \\ = & \left(a_{1}+a_{4}+a_{7}\right)+\left(a_{2}+a_{5}+a_{8}\right)+\left(a_{3}+a_{6}+a_{9}\right) \end{aligned}$$

However, the left side of this equation is the sum of two odd numbers and one even number, while the right side is the sum of 3 odd numbers, leading to a contradiction where the left side is even and the right side is odd.

Therefore, among these 27 sums, there can be at most 24 odd numbers.
The example below (as shown in Figure 2) demonstrates that there exists a way to fill the numbers such that 24 of the 27 sums can be odd. In the tables of Figure 2, 0 represents an even number, and 1 represents an odd number, from left to right, representing the top, middle, and bottom layers of the unit cubes.
Thus, among these 27 sums, the maximum number of odd numbers is 24.