A (192 digit) $123456789101112 \ldots 979899100$ number, delete 100 digits from it so that the remaining digits read together form the largest possible number.
No matter which 100 digits are chosen for deletion, the same number of digits in the written number always remains. To compare two such, we take one digit from each, starting from the front, until we come to two different digits; the number to which the larger of the different digits belongs is the larger, the order of the remaining digits does not affect the magnitude. According to this, the largest number is clearly obtained if we take the largest possible digit for the first, second, ... digit, that is, 9s as long as possible.

The first 9 (before the "10" number) is attainable because it stands in the 9th place. Similarly, the 2nd, 3rd, 4th, 5th 9 (the second digit of the 19, 29, 39, 49 numbers) are also attainable because between two adjacent 9s, 19 other digits stand (of the 20 digits of ten two-digit numbers, the 20th is the next 9), so by keeping only these five 9s, we delete $8+4 \cdot 19=84$ digits. With the remaining 16 deletions, we cannot reach the 9 of the 59 number, nor the next smaller digit, the 8 of the 58, because for that 17 deletions would be needed. The largest attainable digit is the 7 of the 57 number. So far, with 15 more deletions, we have deleted 99 digits. Now with the last deletion, the 8 of the 58 can be reached. Thus, the number in question is $9999978596061 \ldots 99100$.