$9 \cdot 42$ Remove 100 digits from the given number $123456789101112 \cdots 9899100$ so that the remaining number is the largest possible.
[Solution] Since among natural numbers with the same number of digits, the number with a larger first digit is larger, we first remove 84 digits that are not 9 before 50, resulting in 9999950515253545556575899100. Then, starting from the 6th digit of the resulting number, we continuously remove 15 digits $5,0,5,1,5,2,5,3,5,4,5,5,5,6,5$, leaving the 7 that follows. Finally, we remove the 5 in 58. Thus, the resulting number is
$$
999997859606162 \cdots 9899100 \text {, }
$$

It is clear that this number is the one we seek.