4. A figure $\Psi$ composed of 4 unit squares is called a $T$-shaped block, and the figure $\square$ is called an $S$-shaped block. Cover an $8 \times 10$ rectangular chessboard with several $S$-shaped blocks and several $T$-shaped blocks without overlapping, and both shapes can be rotated or flipped. Then, what is the minimum number of $T$-shaped blocks needed?
Analysis: In a tiling method, we categorize the blocks used into the following 4 types: horizontally placed T-shaped blocks (or their flips) are called type $A$, vertically placed T-shaped blocks (or their rotations) are called type $B$, horizontally placed S-shaped blocks (or their flips) are called type $C$, and vertically placed S-shaped blocks (or their flips) are called type $D$. Let the number of these 4 types of blocks be $a, b, c, d$ respectively.

In the $k$-th row of the chessboard, fill each cell with the number $(-1)^{k} k, 1 \leq k \leq 8$, so the sum of all numbers in the table is 40. Each $A$-type block either covers 3 positive numbers and 1 negative number, or 3 negative numbers and 1 positive number, while each $B$-type, $C$-type, and $D$-type block covers 2 positive numbers and 2 negative numbers. Since the number of positive and negative numbers in the entire chessboard is 40 each, $a$ is even, and among the $A$-type blocks, $\frac{a}{2}$ cover 3 positive and 1 negative, and the other $\frac{a}{2}$ cover 3 negative and 1 positive.

For each $A$-type block that covers 3 positive and 1 negative, the sum of the 4 numbers it covers does not exceed $3 \cdot 8 + (-7) = 17$; for each $A$-type block that covers 3 negative and 1 positive, the sum of the 4 numbers it covers does not exceed $3 \cdot (-1) + 2 = -1$. Therefore, the sum of the numbers covered by all $A$-type blocks $\leq 17 \cdot \frac{a}{2} + (-1) \cdot \frac{a}{2} = 8a$.

For each $C$-type block, the sum of the 4 numbers it covers is 2 or -2, so the sum of the numbers covered by all $C$-type blocks $\leq 2c$. Each $B$-type or $D$-type block covers 4 numbers whose sum is always 0. We have $8a + 2c \geq 40$, i.e., $4a + c \geq 20$.

Now, let's fill the numbers in another way. In the $j$-th column of the chessboard, for $1 \leq j \leq 10$, the sum of all numbers in the table is $\mathbf{40}$. Then each $B$-type block either covers 3 positive numbers and 1 negative number, or 3 negative numbers and 1 positive number, while each $A$-type, $C$-type, and $D$-type block covers 2 positive numbers and 2 negative numbers. Since the number of positive and negative numbers in the entire chessboard is 40 each, similarly, $b$ is even, and among the $B$-type blocks, $\frac{b}{2}$ cover 3 positive and 1 negative, and the other $\frac{b}{2}$ cover 3 negative and 1 positive. For each $B$-type block that covers 3 positive and 1 negative, the sum of the 4 numbers it covers does not exceed $3 \cdot 10 + (-9) = 21$; for each $B$-type block that covers 3 negative and 1 positive, the sum of the 4 numbers it covers does not exceed $3 \cdot (-1) + 2 = -1$. Therefore, the sum of the numbers covered by all $B$-type blocks $\leq 21 \cdot \frac{b}{2} + (-1) \cdot \frac{b}{2} = 10b$.

For each $D$-type block, the sum of the 4 numbers it covers is 2 or -2, so the sum of the numbers covered by all $D$-type blocks $\leq 2d$. Each $A$-type or $C$-type block covers 4 numbers whose sum is always 0. We have $10b + 2d \geq 40$, i.e., $5b + d \geq 20$.

By the area relationship, we have $4a + 4b + 4c + 4d = 80$, i.e., $a + b + c + d = 20$. 

From (1) + (2) - (3), we get $3a + 4b \geq 20$, so $4a + 4b \geq 20$, i.e., $a + b \geq 5$. Combining with $a, b$ being even, we have $a + b \geq 6$. The following example shows that the equality can hold, so the number of T-shaped blocks is at least 6.