This is not a solution. It looks like $ 8$ is the maximum number of points. Some trial and error shows tetrahedron does not work for us. Construct a cube with side length of $ 1$ wlog. Clearly we cannot place a point outside the cube. Consider the plane containing $ (0,0,0), (0,0,1),(1,1,1)$ and the other perpendicular to it that contains the other body diagonal. A plane partitions the interior space of the cube. Any point placed in one of the partition makes obtuse angle with a pair of points on the body diagonal. Since we cannot place the point at the centre of the cube, we cannot place a point in the interior of the cube.

Tetrahedron and cube are the only two regular convex polyhedra with internal angle between two faces $ \leq \frac{\pi}{2}$. Other than the four vertices of the tetrahedron we cannot place a point ouside or inside or on the surface of the tetrahedron. Similarly we cannot place a point (other than vertices) inside or outside or on the surface of the cube. Hence we can place at most $ 8$ points with the given restriction.

First, given a regular tetrahedron there most certainly are points that I can add to create no obtuse triangles: each regular tetrahedron contains 4 vertices of a cube, and we can just add the other 4. Second, that argument is worthless: just because you can't add another point once you've already chosen 8 points that are vertices of a cube definitely does not imply that you can't choose 9 points.

JBL wrote: just because you can't add another point once you've already chosen 8 points that are vertices of a cube definitely does not imply that you can't choose 9 points. It is clear that we cannot add a point once the cube is constructed. If we did have a polyhedron with greater than $ 8$ vertices, then at least one of the internal angle exceeds $ \frac{\pi}{2}$.(A regular convex polyhedron with more than $ 8$ vertices has internal angle between any two touching planes $ \geq \frac{\pi}{2}$). Fine ?

JBL wrote: each regular tetrahedron contains 4 vertices of a cube. Um, I think there's some typo or something here.

I meant, "For any regular tetrahedron there exists a cube such that the vertices of the tetrahedron are a subset of the vertices of the cube." In particular, if you have four points at the vertices of a regular tetrahedron, it is possible to add at least four additional points without violating the given condition.

JBL wrote: "For any regular tetrahedron there exists a cube such that the vertices of the tetrahedron are a subset of the vertices of the cube." Wow. I never knew that. {(0,0,0),(0,1,1),(1,0,1),(1,1,0)} is a tetrahedron. Never thought about it. Thanks JBL. I'm sorry.

Well...this is actually a well-know problem. The $d$ dimensional case was conjuctured by Erdos around $1950$ and solved by Ludwig Danzer and Branko Grunbaum in 1962. You may check further infomation from $Proofs$ $from$ $the$ $book$!