Let our tetrahedron be $ABCD$, and let $A_1, A_2, A_3, A_4, A_5, A_6$ denote the midpoints of $AB, AC, AD, CD, BD, BC$, respectively. Let $a, b, c, d$ be vectors on the unit sphere corresponding to $A, B, C, D$, respectively. We claim that the center of the sphere containing the midpoints is $\frac{a+b+c+d}{4}.$ Indeed, firstly observe that this center $O$ must be unique, say it corresponds to the vector $o$. Then, if we consider the point $O'$ which corresponds to the vector $o' = \frac{a+b+c+d}{2} - o$, it's easy to see that it is also equidistant to all the $A_i$'s ($A_1OA_4O'$, etc. are all kites). Therefore, by uniqueness, we have that $o = \frac{a+b+c+d}{2} - o \Rightarrow o = \frac{a+b+c+d}{4}.$ Notice that this point $O$ is the midpoint of $A_1A_4, A_2A_5, A_3A_6$, so that since all the $A_i$'s are equidistant to $O$, we know that $A_1A_4 = A_2A_5 = A_3A_6.$ We know from Apollonius' Median Formula that $A_1A_4^2 = \frac{AA_4^2 + BA_4^2}{2} - \frac{AB^2}{4} = \frac{\frac{AC^2 + AD^2}{2} - \frac{CD^2}{4} + \frac{BC^2 + BD^2}{2} - \frac{CD^2}{4}}{2} - \frac{AB^2}{4} = \frac{AB^2 + AC^2 + AD^2 + BC^2 + BD^2 + CD^2}{4} - \frac{AB^2 + CD^2}{2}.$ With similar expressions for $A_2A_5^2, A_3A_6^2$, we obtain that $AB^2 + CD^2 = AD^2 + BC^2 = AC^2 + BD^2$. This is easily seen to imply that any two altitudes concur, and so we're done.