Example 6 Consider a $2^{m} \times 2^{m}\left(m \in Z_{+}\right)$ chessboard, which is divided into several rectangles composed of the squares of the chessboard, and each of the $2^{m}$ squares on one of the diagonals is a rectangle of side length 1. Find the minimum value of the sum of the perimeters of the divided rectangles. ${ }^{[4]}$
Since $(i, i)$ is separated out, the main diagonal divides the chessboard into two symmetrical parts, and we only need to discuss the lower part.

Let $R_{i}$ denote the set of rectangles that occupy at least one square in the $i$-th row, and $C_{i}$ denote the set of rectangles that occupy at least one square in the $i$-th column. Then the total perimeter of all these rectangles is $P=2 \sum\left(\left|R_{i}\right|+\left|C_{i}\right|\right)$, because each rectangle in $R_{i}$ occupies two vertical line segments in the $i$-th row, and each rectangle in $C_{i}$ occupies two horizontal line segments in the $i$-th column.

Now, randomly select some rectangles from this partition with a probability of $\frac{1}{2}$ to form a set $S$. Let $E_{i}$ be the event that all rectangles in $R_{i}$ are selected and no rectangles in $C_{i}$ are selected.

Clearly, $P\left(E_{i}\right)=2^{-\left(\left|R_{i}\right|+\left|c_{i}\right|\right)}$. For different $i$ and $j$, event $E_{i}$ indicates that the rectangles in set $S$ cover $(i, j)$, while event $E_{j}$ indicates that the rectangles in set $S$ do not cover $(i, j)$. Therefore, $E_{i}$ and $E_{j}$ cannot occur simultaneously.
Thus, $\sum_{i=1}^{2^{m}} 2^{-\left(\left|R_{i}\right|+\left|c_{i}\right|\right)} \leqslant 1$.
By the mean inequality, we have
$$
\begin{array}{l}
\sum_{i=1}^{2^{m}} 2^{-\left(\left|R_{i}\right|+\left|c_{i}\right|\right)} \geqslant 2^{m} \times 2^{-\sum_{i=1}^{m^{m}} \frac{\left|R_{i}\right|+\left|c_{i}\right|}{2^{m}}} \\
=2^{m} \times 2^{-\frac{p}{2^{m+1}}} \\
\Rightarrow 2^{\frac{p}{2^{m+1}}} \geqslant 2^{m} \Rightarrow P \geqslant 2^{m+1} m .
\end{array}
$$

Similarly, the perimeter of the upper part is at least $P$, plus the $2^{m}$ blocks on the main diagonal, the total perimeter is at least
$$
2^{m+2}(m+1) \text {. }
$$

Below, we provide an example to show that such a partition exists.
Use recursive construction.
For $m=1$, it is clearly true. Divide the $2^{m} \times 2^{m}$ into four $2^{m-1} \times 2^{m-1}$, and the two that are crossed by the main diagonal, using the recursive construction, the perimeter is at least $2^{m+1} m \times 2=2^{m+2} m$, the other two blocks are not divided, the perimeter is the perimeter of a square $2 \times 4 \times 2^{m-1}=2^{m+2}$, adding them together gives the answer.