I see two of this year's Ibero problems were proposed by you. Did you propose this one too?

No, I didn't propose the problem and I haven't heard who did I found this problem pretty challenging and I didn't manage to solve it when I tried Ibero day 1 for a few hours. For the sake of having a complete solution of this problem on aops, here is one based on ideas communicated to me by a friend that attended Ibero. Sort the lines by their slope. This is defined as no two lines are parallel. We claim that the "middle" line according to this sorting works if we choose $P$ to be a point on the line which leaves all of the intersections if the chosen line with another line on the same side (if $n$ is even then both middle lines work) . In what follows we define everything with respect to this line unless specified otherwise. We define the level of a point on one of the lines to be the number of lines that are "below" the chosen line (formally, if we take an open ray starting at that point that goes in the negative $y$ direction, then the level ks the number of lines that are intersected). Define a path to be the set of points visited by a point $P$ if we extend the process described in the problem infinitely in both directions. Claim. The different paths partition the points in the $n$ lines into different levels. Proof. It is straightforward to verify that every point in the configuration belongs to some path, so it suffices to show that every path contains only points from the same level. Notice that the level of a poimt travelling along a path could only possibly change when a line $\ell$ is intersected by a line $\ell'$ with a smaller slope. But in this case our rule dictates that we will continue along $\ell'$ in such a way that $\ell$ is now below the moving point, so the level does not change. Now, by our choice of $P$, the line it is on has level $\left\lfloor \frac{n} {2} \right \rfloor$, so it suffices to show that every line contains a point of this level. To this effect fix one of the lines and choose points $P_1$ and $P_2$ on the line kn such a way that every intersection point of the fixed line with another is contained in the segment determined by $P_1$ and $P_2$. Then it's easy to verify that the levels of $P_1$ and $P_2$ add up tp $n - 1$, because every line that lies below $P_1$ must lie above $P_2$ and vice versa. So one of the levels is greater than $\left\lfloor \frac{n} {2}\right \rfloor$ and the other one is less than that. But because no three lines are concurrent, if we move from $P_1$ to $P_2$ along the fixed line, the level only changes by unit amounts whenever it does change. It follows that at some point we had a point with the desired level.

Is this problem some kind of dual version of the Windmill Problem IMO 2011 P2?

ZeusDM wrote: Is this problem some kind of dual version of the Windmill Problem IMO 2011 P2? Yes, yes it is. Fix an arbitrary circle $\omega$ and consider the pole of each line and the polar of $P$ wrt $\omega$. This is clearly equivalent to IMO 2011/2. Choose $\omega$ so that the polar of $P$ passes through the center of $\omega$ after having touched every point. As the polar of $P$ passes through the origin, it will have hit every point already. This corresponds to $P$ moving arbitrarily far away after having been on every line. Remark: This solution also implies that this process cannot continue indefinitely, since the polar of $P$ must eventually pass through the center of $\omega$.

I see this problem and the first thing I think of is graphs. Is there any way to solve it with graph theory?