1-195 1 to 1982 natural numbers are arranged in a certain order, one after another. The computer reads two adjacent numbers from left to right (the 1st and the 2nd, the 2nd and the 3rd, etc.), until the last two numbers, and if the larger number is on the left, the computer swaps their positions. Then the computer reads from right to left in the same way and swaps the positions of two numbers according to the same rule. After reading, the information is obtained: the number in the 100th position did not change its position in both readings. Find this number.
[Solution] Let the largest number among the first 99 numbers be $a$, the smallest number among the last 1882 numbers be $b$, and the 100th number be $c$. If $a>c$, then when the computer reads from left to right, it will move the position of $a$ to the left of $c$, causing the number $c$ to change its position, which contradicts the problem statement. Therefore, $a \leq c$. Since $b \geq c$, it follows that all the 1882 numbers after $c$ are greater than $c$. Among the numbers from 1 to 1982, only the number 100 can have 99 numbers smaller than it and 1882 numbers larger than it. Hence, $c=100$.