Pick four lattice points with the same norm and overall GCD=1 and make a tetrahedron.
In my excursion into this problem, the norm always turned out to be of the form $3 (2 n-1)^2$. Are all solutions of that form? Here are a few of the normed tetrahedra:
{{{1, 1, 1}, {1, -1, -1}, {-1, 1, -1}, {-1, -1, 1}},
{{3, 3, 3}, {-5, 1, 1}, {1, -5, 1}, {1, 1, -5}},
{{1, 5, 7}, {-1, 5, -7}, {-7, -5, 1}, {7, -5, -1}},
{{7, 7, 7}, {-11, 5, -1}, {-1, -11, 5}, {5, -1, -11}},
{{1, 11, 11}, {7, -13, 5}, {7, 5, -13}, {-15, -3, -3}},
{{1, 1, 19}, {-5, 17, -7}, {17, -5, -7}, {-13, -13, -5}},
{{7, 13, 17}, {-7, 13, -17}, {-17, -13, 7}, {17, -13, -7}},
{{13, 13, 13}, {-5, -19, 11}, {11, -5, -19}, {-19, 11, -5}},
{{1, 7, 25}, {3, 21, -15}, {-23, -11, -5}, {19, -17, -5}},
{{7, 17, 23}, {-23, -17, 7}, {-7, 17, -23}, {23, -17, -7}},
{{1, 5, 29}, {1, -29, -5}, {-25, 11, -11}, {23, 13, -13}},
{{5, 23, 23}, {29, -11, -11}, {-17, 13, -25}, {-17, -25, 13}},
{{19, 19, 19}, {1, -31, 11}, {11, 1, -31}, {-31, 11, 1}},
{{21, 21, 21}, {-11, -29, 19}, {19, -11, -29}, {-29, 19, -11}},
{{1, 19, 35}, {25, -31, 1}, {-37, -13, -7}, {11, 25, -29}},
{{17, 25, 31}, {-31, -25, 17}, {-17, 25, -31}, {31, -25, -17}},
{{1, 5, 43}, {17, 35, -19}, {23, -35, -11}, {-41, -5, -13}},
{{3, 33, 33}, {43, -13, -13}, {-23, 17, -37}, {-23, -37, 17}},
{{1, 31, 35}, {43, -17, -7}, {-29, -35, 11}, {-15, 21, -39}},
{{1, 29, 41}, {-41, -29, 1}, {-1, 29, -41}, {41, -29, -1}},
{{1, 11, 49}, {47, -5, -17}, {-25, -43, -7}, {-23, 37, -25}},
{{31, 31, 31}, {-19, -41, 29}, {29, -19, -41}, {-41, 29, -19}},
{{5, 7, 53}, {7, 47, -25}, {-49, -19, -11}, {37, -35, -17}},
{{15, 39, 39}, {47, -23, -23}, {-31, 25, -41}, {-31, -41, 25}},
{{5, 29, 49}, {-37, -43, 7}, {51, -21, -15}, {-19, 35, -41}},
{{3, 3, 57}, {13, -53, -17}, {-53, 13, -17}, {37, 37, -23}},
{{7, 35, 49}, {53, -5, -29}, {-19, -55, 17}, {-41, 25, -37}},
{{7, 35, 49}, {23, -55, 11}, {29, 25, -47}, {-59, -5, -13}},
{{23, 37, 47}, {-47, -37, 23}, {-23, 37, -47}, {47, -37, -23}},
{{37, 37, 37}, {5, -61, 19}, {19, 5, -61}, {-61, 19, 5}},
{{1, 25, 59}, {-49, -41, -5}, {-7, 47, -43}, {55, -31, -11}},
{{39, 39, 39}, {-59, -11, 31}, {31, -59, -11}, {-11, 31, -59}},
{{1, 29, 61}, {-5, -67, 7}, {57, 15, -33}, {-53, 23, -35}}
}
Yes, they are, and here is why.
First we find two integer vectors of equal length forming the correct tetrahedral angle, which is $\arccos\left(-{1\over3}\right)$. Suppose we did that; how do we find the other two vectors? Why, that's simple: find the half-sum of the existing two (that would be our axis), rotate them around the axis by $90^\circ$, and invert.
The general form of rotation matrix by the angle $\alpha$ around the vector $(x,y,z)$ with norm $1$ is $$\left(\begin{array}{c|c|c} x^2+(1-x^2)\cos\alpha & xy(1-\cos\alpha)+z\sin\alpha& xz(1-\cos\alpha)-y\sin\alpha\\ \hline xy(1-\cos\alpha)-z\sin\alpha& y^2+(1-y^2)\cos\alpha & yz(1-\cos\alpha)+x\sin\alpha\\ \hline xz(1-\cos\alpha)+y\sin\alpha& yz(1-\cos\alpha)-x\sin\alpha & z^2+(1-z^2)\cos\alpha \\ \end{array}\right)$$
With $\alpha=90^\circ$, this simplifies to $$\left(\begin{array}{ccc} x^2 & xy+z& xz-y\\ xy-z& y^2 & yz+x\\ xz+y& yz-x & z^2 \\ \end{array}\right)$$
Now what is our $(x,y,z)$? It is the said half-sum, divided by its own length ($\ell$). What is its squared length ($\ell^2$)? It is $1\over3$ of that of the original two vectors (given that they have equal length, and their angle is what it is), and hence a rational number. Can $\ell$ be irrational? No, because then the terms like $xy+z$ are all irrational, which pretty much (*) ruins the chances of getting an integer vector when acting with that matrix upon another integer vector.
(*) OK, there is a subtle moment: what if we find some integer vector $(a,b,c)$ such that the irrational terms negate each other? Why, let's see: the interplay between $xy+z$ and $xz+y$ gives us $b\over c$, the other terms fix $a\over b$ and (redundantly) $a\over c$, so in fact the mysterious vector is defined uniquely up to collinearity, which means that both original vectors (not being collinear) cannot act like that at the same time.
So $\ell$ is rational; also, $3\ell^2$ is an integer, which leaves for $\ell$ no other way than to be an integer as well. By looking at the things $\mod 4$, we see that $\ell$ must either be odd, or the original vectors' components must all be even. The latter option is ruled out by the condition of overall GCD=1.