Three. (Full marks 25 points) Does there exist 100 different lines in the plane, such that their intersection points total 1998?
Three, there are 1998 intersection points.
First, divide 98 lines into two sets of parallel lines, one set has $k$ lines, and the other set has $(98-k)$ lines. The number of intersection points is $k(98-k)$. When the value of the positive integer $1998-k(98-k)$ is minimized, $k=26$ or 72.

When $k=26$ or 72, $k(98-k)=26 \times 72=1872$ intersection points. This leaves $1998-1872=126$ more intersection points needed.

Now add two parallel lines, and construct them appropriately so that they intersect the first 98 lines at 126 points different from the initial 1872 points. The specific construction is as follows:

In the coordinate system xoy, take the lines $x=0, x=1$, $x=2, \cdots, x=71, y=0, y=1, y=2, \cdots, y=25$. These 98 lines have 1872 intersection points.

The 99th line $y=x+8$ intersects the first 98 lines at 80 points: $A_{0}(-8,0), A_{1}(-7,1), A_{2}(-6,2), \cdots$, $A_{7}(-1,7), A_{8}(0,8), A_{9}(1,9), A_{10}(2,10), \cdots, A_{25}(17,25), A_{25}(18,26), \cdots, A_{79}(71,79)$, where points $A_{8}$, $A_{9}, A_{10}, \cdots, A_{25}$, these 18 points are among the initial 1872 intersection points, i.e., they are repeated intersection points.

The 100th line, parallel to the 99th line, $y=x+9$ intersects the first 98 lines at 81 points: $B_{0}(-9,0), B_{1}(-8, 1), B_{2}(-7,2), \cdots, B_{8}(-1,8), B_{9}(0,9), B_{10}(1,10)$, $\cdots, B_{25}(16,25), B_{25}(17,26), B_{27}(18,27), \cdots, B_{80}(71, 80)$. Points $B_{9}, B_{10}, B_{11}, \cdots, B_{25}$, these 17 points are among the initial 1872 intersection points, i.e., they are repeated intersection points (as shown in the figure).
Thus, the number of intersection points of these 100 lines is:
$$
1872+(80-18)+(81-17)=1998(\text{ points }).
$$