Example 12. Is there a north-facing 100 petals arrangement, such that they exactly have 1985 intersections.
Solution 1: Since $x$ parallel lines and another set of $100 - x$ parallel lines can yield $x(100-x)$ intersection points, and
$$
x(100-x)=1985
$$

has no integer solutions, we can consider a grid formed by 99 lines. Since
$$
x(99-x)<1955
$$

has solutions $x \leqslant 26$ or $x \geqslant 73, x \in \mathbb{N}$, and
$$
1985=73 \times 20+90-12,
$$

we can make the following selection:
(1) 73 horizontal lines and 26 vertical lines
$$
\begin{array}{l}
x=k, k=1,2, \cdots, 73, \\
y=k, k=1,2, \cdots, 26 .
\end{array}
$$

These 99 lines can yield $73 \times 26$ intersection points.
(2) Draw the line $y=x+14$ intersecting with the above 99 lines, yielding 99 intersection points, among which 12 points $(1,15),(2,15), \cdots,(12,26)$ are excluded, leaving $99-12$ intersection points.

From (1) and (2), 100 lines can yield $73 \times 26 + 99 - 12 = 1985$ intersection points.

Solution 2: Among 100 lines, if no two lines are parallel and no three lines are concurrent, then there can be $C_{100}^{2}=4950$ intersection points. Now, using concurrent lines to reduce the number of intersection points, each set of $n$ concurrent lines reduces the number of intersection points by $C_{n}^{2}-1$.

Suppose there are $k$ sets of concurrent lines, with the number of lines in each set being $n_1, n_2, \cdots, n_k$. Then we have
$$
\begin{array}{l}
n_1 + n_2 + \cdots + n_k \leqslant 100, \\
C_{n_1}^{2}-1 + C_{n_2}^{2}-1 + \cdots + C_{n_k}^{2}-1 = 2965 \text{ (group 1) } \\
C_{100}^{2} - 2965 \text{ (group 2) }.
\end{array}
$$

To satisfy
$$
C_{n_1}^{2}-1 < 2965,
$$

the largest integer $n_1 = 77$, at which point
$$
C_{77}^{2}-1 = 2925.
$$

Therefore, we can construct a set of 77 concurrent lines, which still needs to reduce 40 intersection points. To satisfy
$$
C_{n_2}^{2}-1 < 40,
$$

the largest integer $n_2 = 9$, at which point
$$
C_{9}^{2}-1 = 35.
$$

Therefore, we can construct another set of 9 concurrent lines, which still needs to reduce 5 intersection points.

Since $C_{4}^{2}-1 = 5$, we can construct a set of 4 concurrent lines.

Since $77 + 9 + 4 = 90 < 100$, the 100 lines can be constructed as 77, 9, and 4 concurrent lines, with the remaining 10 lines left unchanged.