 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 givesThe hole dispersion with ${\mathbf{k}}_h=-\mathbf{k}$ gives ${\mathbf{v}}_h({\mathbf{k}}_h)={\mathbf{v}}_e(\mathbf{k})$, soLikewise $\hslash \dot{\mathbf{k}}=-e\mathbf{E}$ becomes $\hslash {\dot{\mathbf{k}}}_h=+e\mathbf{E}$. 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.