If this existed, would we have a tetrahedron in space with odd integer sidelengths and volume $ 0$. I will show this is not possible. Take one of the points as the origin and label the vectors to the other three points $ x,y,z$, then $ V = |x\cdot(y\times z)|$. So $ V^2 = \left|\begin{array}{ccc}x\cdot x & x\cdot y & x\cdot z \\ y\cdot x & y\cdot y & y\cdot z \\ z\cdot x & z\cdot y & z\cdot z\end{array}\right|$, and since $ 2a\cdot b = |a|^2 + |b|^2 - |a - b|^2$ we can express this as $ 8V^2 = \left|\begin{array}{ccc}2|a|^2 & |a|^2 + |b|^2 - |a - b|^2 & |a|^2 + |c|^2 + |a - c|^2 \\ |a|^2 + |b|^2 - |a - b|^2 & 2|b|^2 & |b|^2 + |c|^2 - |b - c|^2 \\ |a|^2 + |c|^2 + |a - c|^2 & |b|^2 + |c|^2 - |b - c|^2 & 2|c|^2 \end{array}\right|$. But if we assume that $ |a|,|b|,|c|,|a - b|,|a - c|,|b - c|$ are odd integers, then of course (squares are 1 mod 8) Sarrus' rule gives $ 8V^2\equiv 4\pmod 8$, so $ V\not = 0$, contradiction.

Peter, for your formula for Volume, shouldn't the triple scalar product be divided by 6?

Yes, it does, however it is irrelevant here, since it only adds a factor of $36$, and the conclusion stays.