Iberoamerican Mathematical Olympiad

Solution:

Let the points be $P_i$ for $i = 1, 2, \ldots, 1997$. Take $P_1$ to be the center of the given unit circle. Let $x_i$ be the distance from $P_i$ to the closest of the other $1996$ points. Let $C_i$ be the circle centered at $P_i$ with radius $x_i / 2$. Then $C_i$ and $C_j$ cannot overlap by more than one point because $x_i$ and $x_j \leq P_i P_j$. Also $x_i \leq 1$, since $P_1 P_i \leq 1$. Thus $C_i$ is entirely contained in the circle centered at $P_1$ with radius $3/2$. Since the circles $C_i$ do not overlap, their total area cannot exceed the area of the circle of radius $3/2$. Hence
$$\frac{x_1^2 + x_2^2 + \ldots + x_{1997}^2}{4} \leq \frac{9}{4}$$
which gives
$$x_1^2 + x_2^2 + \ldots + x_{1997}^2 \leq 9.$$