$11 \cdot 17$ plane, is there 100 different lines that have exactly 1985 different intersection points?
［Solution］There exists. Construct the lines
$$
\begin{array}{l}
l_{j}=\{(x, y) \mid x=j\}, j=0,1, \cdots, 72 ; \\
l_{j}=\{(x, y) \mid y=j-73\}, j=73,74, \cdots, 98,
\end{array}
$$

Then the number of different intersection points among these lines is $73 \times 26=1898$.
Next, select the 100th line such that it intersects with the above 99 lines at exactly 87 new different points. For this, it should intersect with 12 of the 99 lines at existing intersection points and with the remaining 87 lines at new points. Therefore, we choose the 100th line to be the line $x+y=5$. The intersection points $(0,5),(1,4),(2,3),(3,2),(4,1),(5,0)$ are with the 12 lines $\left(l_{j}, j=1,2,3,4,5,6,73,74,75,76,77,78\right)$, and these 6 points have already been counted, while the rest of the intersection points are new. It is evident that these 100 lines have exactly 1985 different intersection points.