Three. (Full marks 25 points) On the blackboard, all natural numbers from 1 to 1997 are written. Students $A$ and $B$ take turns to perform the following operations: Student $A$ subtracts the same natural number from each number on the blackboard (the number subtracted can be different in different operations); Student $B$ erases two numbers from the blackboard and writes down their sum. Student $A$ goes first, and the operations continue until only one number remains on the blackboard. If this number is non-negative, find this number.
Three, because after each operation by student $A$ and student $B$, the number of numbers written on the blackboard decreases by 1. Since students $A$ and $B$ take turns operating, when $B$ completes the last operation, only one number remains on the blackboard, and both have performed 1996 operations. Let $d_{k} (k=1,2, \cdots, 1996)$ be the natural number removed by student $A$ during the $k$-th operation. Since in the $k$-th operation, the sum of the numbers on the blackboard decreases by $(1998-k) d_{k}$, and student $B$'s operation does not change this sum, after 1996 alternating operations, the number written on the blackboard is
\[
\begin{aligned}
x= & (1+2+\cdots+1997)-1997 d_{1}-1996 d_{2}-\cdots \\
& -2 d_{1996} \\
= & 1997\left(1-d_{1}\right)+1996\left(1-d_{2}\right)+\cdots+(1998 \\
& -k)\left(1-d_{k}\right)+\cdots+2\left(1-d_{1996}\right)+1 .
\end{aligned}
\]

For all $k=1,2, \cdots, 1996$, the number $1-d_{k}$ is non-positive.
If for some $k$, $d_{k} \geqslant 2$, then
\[
(1998-k)\left(d_{k}-1\right) \geqslant 2 \text{. }
\]

Thus, $x \leqslant(1998-k)\left(1-d_{k}\right)+1 \leqslant-1$.
This contradicts the conditions given in the problem. Therefore, for all $k=1,2, \cdots, 1996$, $d_{k}=1$. This means $x=1$.
(243000, Anhui Province, Maguan Second Middle School, Cheng Lihu)