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.