Take $v_1,v_2,\dots,v_n$ which span parallelepiped of maximal volume. If $$a_i=x_1\cdot v_1+\dots+x_n\cdot v_n$$ then $|x_k|\le 1$, otherwise exchanging $v_k$ to $a_i$ will increase the volume. Hence $a_i$ belongs to the convex hull of $\{\pm n\cdot v_{i}\}$; i.e. $c=1$.

Below is the original answer to the original question.

If you agree to choose $N=\tfrac{n{\cdot}(n+1)}{2}$ points then you can get $c=1$.

Take the ellipsoid of smallest volume which contains all $\{a_i\}$. You may assume that ${a_i\}$ is in generic position, in this case at most $N$ of the points lie on the boundary of ellipsoid; take them as $\{v_i\}$. If one of $a_i$ does not lie in the convex hull of $\{\pm\sqrt{n}\cdot v_i\}$ then you can decrease the volume of the ellipsoid, by pushing it in one direction and expanding in all the orthogonal directions.

Take $v_1,v_2,\dots,v_n$ which span a parallelepiped of maximal volume. If $$a_i=x_1\cdot v_1+\dots+x_n\cdot v_n$$ then $|x_k|\le 1$, otherwise exchanging $v_k$ to $a_i$ will increase the volume. Hence $a_i$ belongs to the convex hull of $\{\pm n\cdot v_{i}\}$; i.e. $c=1$.

Below is the original answer to the original question.

If you agree to choose $N=\tfrac{n{\cdot}(n+1)}{2}$ points then you can get $c=1$.

Take the ellipsoid of smallest volume which contains all $\{a_i\}$. You may assume that $\{a_i\}$ is in generic position, in this case at most $N$ of the points lie on the boundary of ellipsoid; take them as $\{v_i\}$. If one of $a_i$ does not lie in the convex hull of $\{\pm\sqrt{n}\cdot v_i\}$ then you can decrease the volume of the ellipsoid, by pushing it in one direction and expanding in all the orthogonal directions.

If you agree to choose $N=\tfrac{n{\cdot}(n+1)}{2}$ points then you can get $c=1$.

Take the ellipsoid of smallest volume which contains all $\{a_i\}$. You may assume that ${a_i\}$ is in generic position, in this case at most $N$ of the points lie on the boundary of ellipsoid; take them as $\{v_i\}$. If one of $a_i$ does not lie in the convex hull of $\{\pm\sqrt{n}\cdot v_i\}$ then you can decrease the volume of the ellipsoid, by pushing it in one direction and expanding in all the orthogonal directions.

Take $v_1,v_2,\dots,v_n$ which span parallelepiped of maximal volume. If $$a_i=x_1\cdot v_1+\dots+x_n\cdot v_n$$ then $|x_k|\le 1$, otherwise exchanging $v_k$ to $a_i$ will increase the volume. Hence $a_i$ belongs to the convex hull of $\{\pm n\cdot v_{i}\}$; i.e. $c=1$.

Below is the original answer to the original question.

If you agree to choose $N=\tfrac{n{\cdot}(n+1)}{2}$ points then you can get $c=1$.

Take the ellipsoid of smallest volume which contains all $\{a_i\}$. You may assume that ${a_i\}$ is in generic position, in this case at most $N$ of the points lie on the boundary of ellipsoid; take them as $\{v_i\}$. If one of $a_i$ does not lie in the convex hull of $\{\pm\sqrt{n}\cdot v_i\}$ then you can decrease the volume of the ellipsoid, by pushing it in one direction and expanding in all the orthogonal directions.