Yukihira is counting the minimum number of lines $m$, that can be drawn on the plane so that they intersect in exactly $200$ distinct points.What is $m$?
1. To find the minimum number of lines \( m \) that can be drawn on the plane so that they intersect in exactly 200 distinct points, we start by using the formula for the maximum number of intersections of \( m \) lines. The maximum number of intersections of \( m \) lines is given by:
   \[
   \frac{m(m-1)}{2}
   \]
   This formula comes from the combinatorial selection of 2 lines out of \( m \) lines, which is \( \binom{m}{2} \).

2. We need this maximum number of intersections to be exactly 200. Therefore, we set up the equation:
   \[
   \frac{m(m-1)}{2} = 200
   \]

3. To solve for \( m \), we first multiply both sides of the equation by 2 to clear the fraction:
   \[
   m(m-1) = 400
   \]

4. This simplifies to a quadratic equation:
   \[
   m^2 - m - 400 = 0
   \]

5. We solve this quadratic equation using the quadratic formula \( m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -1 \), and \( c = -400 \):
   \[
   m = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-400)}}{2 \cdot 1}
   \]
   \[
   m = \frac{1 \pm \sqrt{1 + 1600}}{2}
   \]
   \[
   m = \frac{1 \pm \sqrt{1601}}{2}
   \]

6. Calculating the square root and solving for \( m \):
   \[
   \sqrt{1601} \approx 40.0125
   \]
   \[
   m = \frac{1 \pm 40.0125}{2}
   \]

7. We take the positive root since \( m \) must be a positive integer:
   \[
   m = \frac{1 + 40.0125}{2} \approx \frac{41.0125}{2} \approx 20.50625
   \]

8. Since \( m \) must be an integer, we take the ceiling of 20.50625:
   \[
   \lceil 20.50625 \rceil = 21
   \]

9. Therefore, the minimum number of lines required to intersect in exactly 200 distinct points is:
   \[
   m = 21
   \]

The final answer is \( \boxed{21} \)