8.78 On the blackboard, all natural numbers from 1 to 1988 are written. Operations $A$ and $B$ are alternately performed on these numbers, i.e., first $A$, then $B$, then $A$ again, then $B$, and so on. Operation $A$ involves subtracting the same natural number from each number on the blackboard (the number subtracted can be different in different $A$ operations); Operation $B$ involves erasing two numbers from the blackboard and writing down their sum. The process continues until after a certain $B$ operation, only one number remains on the blackboard. Given that this number is non-negative, find this number.
［Solution］Since operation $A$ does not decrease the number of numbers on the blackboard, while operation $B$ reduces the number of numbers by 1 each time, after operation $A$ and operation $B$ are each performed 1987 times, only 1 number remains on the blackboard.

Let $d_{k}$ be the natural number subtracted during the $k$-th operation $A$, where $k=1,2, \cdots$, 1987. Since in the $k$-th operation $A$, the sum of all numbers on the blackboard will decrease by $(1989 - k) d_{k}$, and operation $B$ does not change this sum, after alternating 1987 times of operation $A$ and operation $B$, the number written on the blackboard should be
$$
x=\sum_{k=1}^{1988} k-\sum_{k=1}^{1987}(1989-k) d_{k}=\sum_{k=1}^{1987}(1989-k)\left(1-d_{k}\right)+1 .
$$

Since $d_{k}$ is a natural number, $1-d_{k} \leqslant 0, k=1,2, \cdots, 1987$. Also, since $x \geqslant 0$, it must be that $d_{k}$ $=1, k=1,2, \cdots, 1987$. Thus, we get $x=1$, meaning the last remaining number on the blackboard is 1.