= Solution
Removing an <electron> of charge $-e$ leaves charge $+e$ relative to the filled-band reference. This also reproduces the current. If the missing <electron> state had <velocity> $\mathbf v_e(\mathbf k)$, subtracting its contribution from the zero current of a filled band gives
$$
\Delta\mathbf j=-(-e)\mathbf v_e(\mathbf k)=+e\mathbf v_e(\mathbf k).
$$
The hole dispersion with $\mathbf k_h=-\mathbf k$ gives $\mathbf v_h(\mathbf k_h)=\mathbf v_e(\mathbf k)$, so
$$
\boxed{q_h=+e,\qquad \Delta\mathbf j=q_h\mathbf v_h.}
$$
Likewise $\hbar\dot{\mathbf k}=-e\mathbf E$ becomes $\hbar\dot{\mathbf k}_h=+e\mathbf E$. Positive hole charge and positive hole <effective mass> consistently describe drift along the <electric field>, while the valence <electrons> themselves drift through an almost full band. A hole is a quasiparticle, not a proton moving through the crystal.
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