Problem 982

Answer: It is not possible.
Suppose it is possible to obtain a configuration with exactly one field $P_1$, on which the number $-1$ is written. Due to the standard symmetries of the chessboard $S$, we can assume that the field $P_1$ has its center at $(a-\frac{1}{2},b-\frac{1}{2})$, where $a \geq b \geq 1$ ($a$, $b$ - integers). The point ($a$, $b$) is a vertex of the field $P_1$, so $a^2 + b^2 < 1998^2$.
We will show that the unit square $P_2$ with center $(a-\frac{3}{2},b+\frac{1}{2})$ is also a field of the chessboard $S$.
If $a > b$, then $(a-1)^2 + (b+1)^2 \leq a^2 + b^2 \leq 1998^2$. If, however, $a = b$, then the inequality $a^2 + b^2 = 2a^2 \leq 1998^2$ is definitely not an equality; hence,

Therefore, the point $(a-1,b+1)$, which is the top right vertex of the square $P_2$, lies within the circle mentioned in the problem. Consequently, the entire square $P_2$ is contained within this circle, meaning it is a field of the chessboard $S$.
om49_3r_img_2.jpg
Now consider eight unit squares, two of which are the fields $P_1$ and $P_2$, arranged as shown in the adjacent figure. Since the squares $P_1$ and $P_2$ are fields of the chessboard $S$, the remaining six squares are also fields.
Notice that each vertical, horizontal, or diagonal row either contains exactly two of the eight given fields or none of them. The product of the numbers written on these eight fields does not change as a result of the operations performed and remains constantly equal to $1$. It is therefore not possible for the number $-1$ to appear on the field $P_1$ at some point, while the others have $+1$.