Answer: 23. Solution: Lines $a$ and $b$ have no more than one common point. If such a point exists and is marked, then together on lines $a$ and $b$ there are 79 marked points; otherwise, there are 80. Therefore, outside lines $a$ and $b$, there are no more than 21 marked points.

Take an arbitrary line $c$ that does not coincide with $a$ and $b$. On it, there are no more than 21 points not lying on $a$ and $b$, and, in addition, the points of its intersection with $a$ and $b$ can be marked - a total of no more than 23 points.

Mark the vertices of some triangle $ABC$. On its sides $a=BC$ and $b=AC$, mark another 38 points each, and on side $c=AB$, mark another 21 points. We have constructed an example where exactly 23 points lie on line $c$, which completes the solution.

- See also problem $7-2$.
- For an answer without justification - 0 points. An answer with a correct example but without justification of the estimate - 3 points, the justification of the estimate is evaluated from the remaining 4 points.
Problem 2. On a plane, 100 points are marked. It turns out that on two different lines a and b, there are 40 marked points each. What is the maximum number of marked points that can lie on a line that does not coincide with a and b?