Consider a convex polyhedron without parallel edges and without an edge parallel to any face other than the two faces adjacent to it. Call a pair of points of the polyhedron [i]antipodal[/i] if there exist two parallel planes passing through these points and such that the polyhedron is contained between these planes. Let $A$ be the number of antipodal pairs of vertices, and let $B$ be the number of antipodal pairs of midpoint edges. Determine the difference $A-B$ in terms of the numbers of vertices, edges, and faces.

[i]Proposed by Kei Irei, Japan[/i]
1. **Define the polyhedron and its properties:**
   Let $\mathcal{P}$ denote the convex polyhedron. Let $V$, $E$, and $F$ denote the number of vertices, edges, and faces of $\mathcal{P}$, respectively. Since $\mathcal{P}$ is convex, it satisfies Euler's formula:
   \[
   V - E + F = 2.
   \]

2. **Coloring the sphere:**
   Label the vertices of $\mathcal{P}$ with $1, 2, \ldots, V$. Define a coloring on the unit sphere $\mathcal{S}^2$ as follows: A point $(a, b, c) \in \mathcal{S}^2$ (where $a^2 + b^2 + c^2 = 1$) is colored with the color $i$ if the function $f(x, y, z) = ax + by + cz$ is maximized at vertex $i$. Points can be colored with multiple colors, but the set of such multi-colored points has measure zero on $\mathcal{S}^2$.

3. **Constructing the graph $S$:**
   This coloring defines a simple graph $S$ on $\mathcal{S}^2$. The face number $i$ of $S$ corresponds to the $i$-monochromatic domain of $\mathcal{S}^2$ (the set of all points colored with $i$). The edges of $S$ correspond to boundaries between monochromatic domains, and the vertices of $S$ correspond to intersections of three or more monochromatic domains. Note that $S$ is isomorphic to the dual graph of the graph of $\mathcal{P}$. Furthermore, the edges of $S$ are segments of great circles. Let $V'$, $E'$, and $F'$ denote the number of vertices, edges, and faces of $S$, respectively. We have:
   \[
   V' = F, \quad E' = E, \quad F' = V.
   \]

4. **Reflecting the graph $S$:**
   Take the reflected graph $S'$ on $\mathcal{S}^2$, defined as the reflection of $S$ about the origin $(0, 0, 0)$. Consider the new graph $T$ topologically induced by $S \sqcup S'$. The problem condition implies that edges from $S$ and edges from $S'$ do not intersect except in their interiors; they do not intersect at their endpoints. This also implies that no vertex of $S$ coincides with a vertex of $S'$. Let $k$ denote the number of intersections of an edge from $S$ with an edge from $S'$.

5. **Counting antipodal pairs:**
   Antipodal pairs of vertices of $\mathcal{P}$ correspond to antipodal faces of $T$, and antipodal midpoints of edges of $\mathcal{P}$ correspond to antipodal pairs of intersections of edges from $S$ with edges from $S'$. This means that:
   \[
   2B = k.
   \]
   Furthermore, letting $V''$, $E''$, and $F''$ denote the number of vertices, edges, and faces of $T$, we see that:
   \[
   2A = F''.
   \]

6. **Calculating $V''$, $E''$, and $F''$:**
   Note that:
   \[
   V'' = 2V' + k
   \]
   because $T$ contains the vertices of $S$, the vertices of $S'$, and new vertices formed by intersections of edges from $S$ with edges from $S'$. Furthermore:
   \[
   E'' = 2(E' + k)
   \]
   because each intersection increases $E''$ by 2 (it splits two edges in half). Using Euler's formula for $T$:
   \[
   V'' - E'' + F'' = 2,
   \]
   we calculate:
   \[
   F'' = 2F' + k - 2.
   \]

7. **Finding $A - B$:**
   Thus:
   \[
   A - B = \frac{1}{2}(2A - 2B) = \frac{1}{2}(F'' - k) = \frac{1}{2}(2F' - 2) = F' - 1 = V - 1.
   \]

The final answer is $\boxed{V - 1}$.