  We will show that for any $n\ge N$ points on the $N$-dimensional sphere $S^N$
there is a closed hemisphere which contains at least
$\lfloor\frac{n+N+1}{2}\rfloor$ of these points. This bound is sharp and we
will calculate the amount of sets which realize this value.
  If we change to open hemispheres things will be easier. For any $n$ points on
the sphere there is an open hemisphere which contains at least
$\lfloor\frac{n+1}{2}\rfloor$ of these points, independent of the dimension.
This bound is sharp.
