Draw circles $C_i$ with centers in $P_i$ and radius $\frac{x_i} 2$. It is easy to see that these circles are disjoint. Assume the contrary: $C_i\cap C_j\neq \Phi$. Hence $P_iP_j < \frac{x_i+x_j} 2 \le \max {x_i,x_j }$. Contradiction with the minimality of $x_i$. Since each circle have radius at most $1/2$, because $PP_i\le 1$, all these disjoint sircles are contained in a circle of center P and radius $3/2$. Hence the sum of areas of $C_i$ is less than the area of this circle, hence $\sum \pi\frac{x_i^2} 4\ge\pi \frac 9 4$. The conclusion follows.

Note that the number 1997 have no relevance for the problem.