b) there are no such $n$.

a) A diagonal of the convex $2n$-gon is said to be main if there are $n-1$ vertices of this polygon between the vertices connected by this diagonal. Suppose that there exist a convex $2n$-gon $A_0A_1...A_{2n-1}$ such that its opposite sides are parallel ($A_iA_{i+1} \parallel A_{i+n}A_{i+n+1}$, $i = 0, ..., n-1$, and $A_{2n} = A_n$) and there is no pair of the opposite sides possessing the property described by the problem condition (there exists a straight line that is perpendicular to these sides and intersects each of them). Consider some two opposite sides $A_iA_{i+1}$ and $A_{i+n}A_{i+n+1}$ of such $2n$-gon. Let $O_i$ be the intersection point of the lines $A_iA_{i+n}$ and $A_{i+1}A_{i+n+1}$. Consider the triangles $O_iA_iA_{i+1}$ and $O_iA_{i+n}A_{i+n+1}$ (see Fig. 1).

Fig. 1

Fig. 2

Since $A_iA_{i+1} \parallel A_{i+n}A_{i+n+1}$ we have $\angle O_iA_iA_{i+1} = \angle O_iA_{i+n}A_{i+n+1}$ and $\angle O_iA_{i+n}A_{i+n+1}$ (let $\alpha = \angle O_iA_{i+n}A_{i+1}$ and $\beta = \angle O_iA_{i+n+1}A_{i+n}$). If at least one of these angles is not obtuse, then there exists a straight line $l$ that is perpendicular to the sides $A_iA_{i+1}$, $A_{i+n}A_{i+n+1}$ and intersects each of them. For example, the line passing through $O_i$ perpendicular to the side $A_iA_{i+1}$ can be considered as $l$. Therefore either $\alpha$ or $\beta$ is greater than $90^\circ$.

Any diagonal starting from the vertex of the angle of the $2n$-gon partitions this angle into two angles (we call that these two angles are adjacent to this diagonal). Construct all main diagonals of the $2n$-gon and number (in clockwise direction) all adjacent angles with the numbers from 1 to $4n$ (see Fig. 2). By the above either angle 1 or angle $4n$ is obtuse. Let, without loss of generality, angle 1 is obtuse. Then angle 2 is acute (any inner angle of a convex polygon is less than $180^\circ$). From what has already been proved, it follows that angle 3 is obtuse, and so on. We see that all angles with odd numbers are obtuse, while all angles with even numbers are acute. So one of the adjacent angles is obtuse but the other is acute for any main diagonal. Consider the greatest of the main diagonals (one of them if there are more than one). Let $A_iA_{i+n}$ be such diagonal (see Fig. 1) and let, without loss of generality, $\angle O_iA_iA_{i+1} = \angle O_iA_{i+n}A_{i+n+1} > 90^\circ$. Since in which is impossible since the diagonal $A_iA_{i+n}$ is the greatest one. This contradiction proves that any convex $2n$-gon with pairwise parallel opposite sides has a pair of the opposite sides such that there exists a straight line that is perpendicular to these sides and intersects each of them.

b) Consider a regular $2n$-gon $S$ with the vertices $A_0, A_1, ..., A_{2n-1}$ ($A_{2n} = A_0$). We choose the vector $\vec{v}$ satisfying the relations $\angle(\vec{v}, A_iA_{i+1}) \neq 90^\circ$ for $i \in \{0, ..., 2n-1\}$,

Fig. 3

and $|\angle(\vec{v}, A_{2n-1}A_i)| < 180^\circ/n$, (here $\angle(\vec{v}, \vec{v})$ is the angle between the vectors $\vec{v}$ and $\vec{v}$). For any $\lambda > 0$ we denote by $S_\lambda$ the convex polygon with the vertices $A_0^\lambda, A_1^\lambda, ..., A_{2n-1}^\lambda$, where $A_i^\lambda = A_i + \lambda\vec{v}$ for $i$ from 0 to $n-1$ (from the inequality $|\angle(\vec{v}, A_{2n-1}A_i)| < 180^\circ/n$ it follows that the polygon $S_\lambda$ is convex). In the polygon $S_\lambda$ for $\lambda$ large enough there exists a unique pair $(A_{2n-1}^\lambda, A_{n-1}^\lambda)$ of the opposite sides such that there exists a straight line that is perpendicular to these sides and intersects each of them.