2. Given that among the vertices of a regular 2012-gon, there exist $k$ vertices such that the convex $k$-gon formed by these $k$ vertices has no two parallel sides. Find the maximum value of $k$.


2. The maximum value of $k$ is 1509.

Let $A_{1}, A_{2}, \cdots, A_{2012}$ be the set of vertices of a regular polygon.
Consider the set of four vertices
$$
\begin{array}{l}
\left(A_{1}, A_{2}, A_{1000}, A_{1008}\right),\left(A_{3}, A_{4}, A_{100}, A_{1010}\right), \\
\cdots,\left(A_{1005}, A_{1000}, A_{2011}, A_{2012}\right) .
\end{array}
$$

If $k$ points contain a set of four points
$$
\left(A_{2 i-1}, A_{2 i}, A_{2 i+1005}, A_{2 i+1006}\right),
$$

then the convex $k$-gon obtained has two parallel sides $A_{2 i-1} A_{2 i}$ and $A_{2 i+1005} A_{2 i+1006}$, which is a contradiction.
This indicates that at most three points can be taken from each set of four points, hence
$$
k \leqslant 503 \times 3=1509 \text {. }
$$

Below is an example for $k=1509$.
It is easy to see that in the convex polygon $A_{1} A_{2} \cdots A_{1006} A_{1008} A_{1010} \cdots A_{2012}$, no two sides are parallel.