Consider the convex hull formed by the points. If the number of points on the convex hull is greater than $4$, then some points on the convex hull will form an obtuse angle, so their orthocenter will fall out of the convex hull, contradiction. If the number of points on the convex hull is $4$, then we know that the convex hull would be a rectangle. Let the rectangle be $ABCD$ and WLOG let $AB\geq BC$, consider $AB$ lying on the x-axis, then we take a new point $E$, and let $H$ be the orthocenter of $ABE$, $H'$ be the orthocenter of $CDH$, then $H'$ would have a smaller y-coordinate than $E$ (unless $ABCD$ is a square and $E$ is the center, which does not work). Do this repeatedly and you will get a point falling out of $ABCD$, contradiction. So there must not be any points inside $ABCD$. If the number of points on the convex hull is $3$, then the triangle has to either be right or acute. For an acute triangle you have to add its orthocenter. Call the triangle $ABC$, if any point $P$ is not the orthocenter of $ABC$ but it's in $S$, then consider the orthocenter of $ABP$,$BCP$,$CAP$. At least one of them would fall out of the triangle, so you can't add more points. To conclude, all possibilities are a rectangle (including a square), a right triangle, and a triangle with its orthocenter.