22. (10 points) There are four boxes, $A$, $B$, $C$, and $D$, each containing some small balls. Now, part of the balls in box $A$ are transferred to the other three boxes according to the following requirements: if the box originally had a certain number of balls, the same number of balls is added again. After that, the balls in boxes $B$, $C$, and $D$ are transferred to the other boxes in the same manner. At this point, each of the four boxes has 16 balls. The box that initially contained the most balls is $\qquad$ box, and it contained $\qquad$ balls.
22. (10 points) There are some balls in four boxes $A$, $B$, $C$, and $D$. Now, part of the balls in box $A$ are transferred to the other three boxes according to the following requirements: if the box originally has a certain number of balls, the same number of balls will be added. After that, the balls in boxes $B$, $C$, and $D$ are transferred to the other boxes in the same manner. At this point, each of the four boxes has 16 balls. Which box initially had the most balls, and how many balls did it have?

【Solution】According to the fact that each of the four boxes has 16 balls at the end, the total number of balls is $16 \times 4 = 64$. The last transfer made the number of balls in $A$, $B$, and $C$ double, so before the last transfer, $A$, $B$, and $C$ each had $16 \div 2 = 8$ balls. Since the total number of balls does not change, before the last transfer, $D$ had $64 - 8 - 8 - 8 = 40$ balls. Thus, the situation before the last transfer is: $A 8, B 8, C 8, D 40$;

The second-to-last transfer made the number of balls in $A$, $B$, and $D$ double, so before the second-to-last transfer, $A$ and $B$ each had $8 \div 2 = 4$ balls, and $D$ had $40 \div 2 = 20$ balls. Since the total number of balls does not change, before the second-to-last transfer, $C$ had $64 - 4 - 4 - 20 = 36$ balls. Thus, the situation before the second-to-last transfer is: $A 4, B 4, C 36, D 20$;

Similarly, before the transfer from $B$, $A$ had $4 \div 2 = 2$ balls, $C$ had $36 \div 2 = 18$ balls, and $D$ had $20 \div 2 = 10$ balls. Since the total number of balls does not change, before the transfer from $B$, $B$ had $64 - 2 - 18 - 10 = 34$ balls. Thus, the situation before the transfer from $B$ is: $A 2, B 34, C 18, D 10$;

One more deduction: before the transfer from $A$, $B$ had $34 \div 2 = 17$ balls, $C$ had $18 \div 2 = 9$ balls, and $D$ had $10 \div 2 = 5$ balls. Since the total number of balls does not change, before the transfer from $A$, $A$ had $64 - 17 - 9 - 5 = 33$ balls. Thus, the situation before the transfer from $A$ is: $A 33, B 17, C 9, D 5$;

The situation before the transfer from $A$ is the initial situation, so initially, box $A$ had the most balls, with a total of 33 balls.

Answer: Initially, the box with the most balls is box $A$, which had 33 balls; hence the answer is: $A, 33$.