For the solution see the attached figure here. It is the counterexample with $n=8$. Vertical lines represent all possible vertical lines which have less that $n$ intersection points with given lines.
How I found this counterexample?
By duality of the real projective plane, the problem is equivalent to the following:
Given $n$ distinct points on the plane such that they are not collinear and no two of them lie on the vertical line, is it true that there exists a direction $\mathcal{D}$ such that when passing a line from $\mathcal{D}$ through every given point we obtain $n-1$ or $n-2$ distinct lines?
And the answer is no. The counterexample consists of the vertices of the regular octagon rotated so that no two vertices lie on the vertical line. The figure corresponds to such octagon.
More notes. After the discussion with my colleagues from Katowice we concluded a little more about the problem.
Concerning the equivalent (dual) problem, given a set of $n$ points on the plane, we may consider the set $I$ of all natural numbers $j$ such that there exists a direction $\mathcal{D}$ yielding exactly $j$ distinct lines (when passing a line from $\mathcal{D}$ through every given point).
For a regular polygon with $n=2k$ vertices we have $I=\{k,k+1,2k\}$ and for a regular polygon with $n=2k+1$ vertices we have $I=\{k+1,2k+1\}$. Therefore $n=7$ is sufficient for a counterexample to the original problem.
One more observation is that transforming a regular polygon through an affine bijection of $\mathbb{R}^2$ preserves parallel lines, so the image of $n$ points under such transformation has the same set $I$ as the original points.
