= Solution
Let $\Omega$ be the interior of the <tetrahedron>, and take any constant <vector field> $\mathbf q$. Its <divergence> is zero, so the <divergence theorem> gives
$$
\mathbf q\cdot\sum_{i=1}^4\mathbf V_i=\sum_{i=1}^4\int_{f_i}\mathbf q\cdot\mathbf n_i\,dS=\int_\Omega\nabla\cdot\mathbf q\,dV=0.
$$
Choosing $\mathbf q$ to be each Cartesian basis vector proves $\sum_i\mathbf V_i=\mathbf0$. This is the general fact that <vector areas of a closed polyhedron sum to zero>, specialized to four faces. The outward orientations are essential: the scalar face areas are all positive and do not sum to zero.
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