Pretty little problem! By Euler's formula, we have $E=F+V+2p-2$, where $V$, $E$, $F$ are the number of vertices, edges and faces of the polyhedron, and $2-2p$ is its Euler Characteristic (i.e. $p$ is the number of "holes" in the polyhedron). Now, call $V_b$ the number of vertices that belong to a black face, and let $V_w=V-V_b$. Then, call $E_b$ the number of edges that belong to a black face, and let $E_w=E-E_b$. Obviously, $V_b=E_b$, because if a black face has $k$ black vertices, it has also $k$ black edges, and all black vertices and edges are disjoint. So, we get $E_w=F+V_w+2p-2 \geq F-2$, and we are done.