4. Answer. 190. Solution 1. Among the lines intersecting $y=20-x$, there is the line $y=x$. Moreover, the picture is symmetric with respect to the line $y=x$, so the sum of the abscissas is equal to the sum of the ordinates. Through the point $(19 ; 19)$, $180: 9=20$ lines are drawn, of which 19 intersect the line $y=20-x$. For each point on the line $y=20-x$, the sum of the coordinates is 20, so the sum of the abscissas and ordinates of all points is $20 \cdot 19=380$, then the sum of the abscissas is half of this and equals 190.
Solution 2. Through the point $(19 ; 19)$, $180: 9=20$ lines are drawn, of which 19 intersect the line $y=20-x$. Let the line $y=x$ intersect the line $y=20-x$ at point $A$, then $x_{A}=10$. The remaining 18 lines are divided into pairs, intersecting the line $y=20-x$ at symmetric points $B$ and $C$ (triangles $M A B$ and $M A C$ are equal by a leg and an acute angle). Then $x_{B}+x_{C}=2 x_{A}=20$. The sum of the abscissas of all intersection points is $10+9 \cdot 20=190$.
![](https://cdn.mathpix.com/cropped/2024_05_06_1b13eede35121da4a712g-3.jpg?height=1148&width=510&top_left_y=313&top_left_x=1412)

Grading criteria. Correct solution - 7 points. An error in calculating the number of lines passing through the point (19;19): 360:9=40, all other reasoning is correct - 4 points. It is justified that the picture is symmetric with respect to the line $y=x$, but there is no further progress - 2 points. In all other cases - 0 points.