$7 \cdot 89$ On 400 cards, the natural numbers $1,2,3, \cdots, 400$ are written. A and B play the following game: In the first step, A arbitrarily takes 200 cards for himself. B then takes 100 cards each from the remaining 200 cards and the 200 cards in A's hand for himself, leaving the remaining 200 cards for A. In the next step, A takes 100 cards each from the cards in both of their hands for himself, leaving the remaining 200 cards for B. This continues until B completes the 200th step, after which both calculate the sum of the numbers on the cards in their hands, $C_{A}, C_{B}$, and then A pays B the difference $C_{B}-C_{A}$. What is the maximum difference B can obtain if both play with the correct strategy?
[Solution] Suppose that before the last card selection by $B$, the numbers on the 200 cards in $A$'s hand are $x_{1}>x_{2}>x_{3}>\cdots>x_{200}$, and the numbers on the 200 cards in $B$'s hand are $y_{1}>y_{2}>y_{3}>\cdots>y_{200}$. Thus, when $B$ makes a move, $B$ can choose $x_{1}, x_{2}, \cdots, x_{100}$ and $y_{1}, y_{2}, \cdots, y_{100}$, so
$$
C_{B}-C_{A}=\sum_{i=1}^{100}\left[\left(x_{i}-x_{100+i}\right)+\left(y_{i}-y_{100+i}\right)\right] \geqslant 20000 .
$$

On the other hand, $A$ selects the numbers $1,2, \cdots, 200$ for themselves in the first step and gives $201,202, \cdots, 400$ to $B$, and each time $A$ makes a move, they always return these numbers to their original state. In this case, the maximum difference $B$ can achieve after making a move is exactly 20000.