1-191 Write down all natural numbers from 1 to 1988 on a blackboard. Perform operations $A$ and $B$ repeatedly on these numbers: first $A$, then $B$, followed by $A$ again, and then $B$, and so on. Operation $A$ involves subtracting the same natural number from each number written on the blackboard (the subtrahend can be different for different instances of operation $A$). Operation $B$ involves erasing two numbers from the blackboard and writing down their sum. Operations $A$ and $B$ are performed in this sequence until, after a certain operation $B$, only one number remains on the blackboard, and it is non-negative. What is this number?
[Solution] Each time operations $A$ and $B$ are performed once, one number is erased from the blackboard. Therefore, after performing operations $A$ and $B$ 1987 times each, only one number remains on the blackboard.

Let the natural number $d_{k}$ be the subtrahend when the $k$-th operation $A$ is performed, where $k=1,2, \cdots, 1987$. After the $k$-th operation $A$, the sum of the numbers written on the blackboard decreases by $(1989-k) d_{k}$; while after operation $B$, this sum remains unchanged. After performing operations $A$ and $B$ 1987 times each, the number written on the blackboard is
$$\begin{aligned}
x= & (1+2+\cdots+1988)-1988 d_{1}-1987 d_{2}-\cdots-2 d_{1987} \\
= & 1988\left(1-d_{1}\right)+1987\left(1-d_{2}\right)+\cdots+(1989-k)\left(1-d_{k}\right) \\
& +\cdots+2\left(1-d_{1987}\right)+1 .
\end{aligned}$$

For all $k=1,2, \cdots, 1987$,
$$(1989-k)\left(1-d_{k}\right) \leqslant 0$$

If for some $k$, $d_{k} \geqslant 2$, then
$$(1989-k)\left(1-d_{k}\right) \leqslant-2$$

Thus, we have
$$x \leqslant(1989-k)\left(1-d_{k}\right)+1 \leqslant-1$$

This contradicts the problem statement. Therefore, for all $k=1,2, \cdots, 1987$, we have $d_{k}=1$, and thus $x=1$.

That is, the number left on the blackboard is 1.