Problem 324

We know that $2008=24 \cdot 72 + 99 - 4 + 99 - 6 + 99 - 7$. Consider the 99 lines $y=1, y=2, \ldots, y=24$ and $x=1, x=2, \ldots, x=72$. They intersect at $72 \cdot 24 = 1728$ points. The line $y=x+20$ intersects the previous lines, but the points $(1,21),(1,22),(1,23)$ and $(1,24)$ have already been counted in the product $72 \cdot 24$. This line thus adds $99-4$ intersection points. Similarly, the lines $y=x+17$ and $y=x+16$ add $99-6$ and $99-7$ new intersection points, respectively. Therefore, we have indeed constructed 100 lines with 2008 intersection points.