4. Fold $\triangle D_{1} D_{2} D_{3}$ along its three medians to form tetrahedron $A B C D$, where, $D_{1} D_{2}=24, D_{1} D_{3}=20$, $D_{2} D_{3}=16$. Then the volume of tetrahedron $A B C D$ is equal to
4. $30 \sqrt{6}$.

The folded tetrahedron has equal opposite edges, with the three pairs of opposite edges being $12$, $10$, and $8$. Therefore, the tetrahedron can be placed in a rectangular box with dimensions $x$, $y$, and $z$, corresponding to the six face diagonals. Thus,
$$
\left\{\begin{array} { l } 
{ x ^ { 2 } + y ^ { 2 } = 1 4 4 , } \\
{ y ^ { 2 } + z ^ { 2 } = 6 4 , } \\
{ x ^ { 2 } + z ^ { 2 } = 1 0 0 }
\end{array} \Rightarrow \left\{\begin{array}{l}
x^{2}=90, \\
y^{2}=54, \\
z^{2}=10 .
\end{array}\right.\right.
$$

Therefore, $V=x y z-4 \times \frac{1}{6} x y z=\frac{1}{3} x y z=30 \sqrt{6}$.