1. The natural numbers from 1951 to 1982 are arranged in a certain order, one after another. The computer reads each pair of adjacent numbers from left to right (the 1st and 2nd, the 2nd and 3rd, and so on), up to the last two numbers. If the larger number is on the left, the computer swaps their positions. Then, the computer reads the numbers from right to left in the same manner, and swaps the positions of the numbers according to the same rule. After reading, it is known that the number in the 100th position did not change its position in either direction. 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 the last 1882 numbers are all 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, $c=100$.