Given a sphere, a great circle of the sphere is a circle on the sphere whose diameter is also a diameter of the sphere. For a given positive integer $n,$ the surface of a sphere is divided into several regions by $n$ great circles, and each region is colored black or white. We say that a coloring is good if any two adjacent regions (that share an arc as boundary, not just a finite number of points) have different colors. Find, with proof, all positive integers $n$ such that in every good coloring with $n$ great circles, the sum of the areas of the black regions is equal to the sum of the areas of the white regions.
1. **Understanding the Problem:**
   We need to determine for which positive integers \( n \) the surface of a sphere divided by \( n \) great circles can be colored such that the sum of the areas of the black regions equals the sum of the areas of the white regions, given that adjacent regions must have different colors.

2. **Initial Claim:**
   We claim that the answer is \( \boxed{n \text{ odd}} \).

3. **Homothety Argument:**
   For every region \(\mathcal{R}\) on the surface of the sphere, there is another region on the sphere that can be mapped to \(\mathcal{R}\) by a homothety \(\mathcal{H}\) of ratio \(-1\) through the center of the sphere.

   - Consider an edge \( e \) of a region formed by two great circles. This edge has a length equal to half the circumference of the great circle.
   - A homothety of \(-1\) through the center of the sphere maps this edge onto another identical edge.
   - Similarly, all intersections between great circles can be mapped through the same homothety to another intersection.
   - Thus, for every region \(\mathcal{R}\), we can map all the edges and vertices via \(\mathcal{H}\) to another identical region.

4. **Path Length Argument:**
   Consider a region \( F \) and let the identical region opposite it on the sphere be \( F' \).

   - **Claim:** Any path of minimum length, where we go from adjacent face to adjacent face, from \( F \) to \( F' \) must have length \( n+1 \) (including \( F \) and \( F' \) themselves).

   - **Proof:**
     - For every great circle \( C \), the region \( F' \) is on the other side of \( C \) than \( F \) is.
     - Assume for contradiction that there exists a great circle \( C \) such that \( F' \) and \( F \) are on the same side. Then a homothety of \(-1\) centered at the center of the sphere takes \( F \) to the other side from itself, but \( F' \) is not on that side, leading to a contradiction.
     - Any path from \( F \) which crosses each great circle exactly once must land at \( F' \). Assume it lands at region \( X \neq F' \). Then there must exist at least one great circle on which \( X \) and \( F' \) are on opposite sides. But that implies that \( X \) is on the same side of said great circle as \( F \), which is impossible.
     - Thus, from \( F \), it can cross over one of the great circles that bound \( F \) into another region \( R_1 \), and from \( R_1 \), it can cross over any great circle not yet crossed into \( R_2 \), and so on until it reaches \( F' \). Since it crosses \( n \) great circles, that path consists of \( n+1 \) regions.

5. **Odd \( n \) Case:**
   - For odd \( n \), consider the aforementioned minimal length path. Then, it must alternate between black and white faces, so \( F \) and \( F' \) will be different colors.
   - Thus, the sum of the black and white areas will be the same.

6. **Even \( n \) Case:**
   - For even \( n \), consider the following construction:
     - Call the \( n \) great circles \( c_1, c_2, \dots, c_n \).
     - Assume they all share exactly one diameter.
     - Assume that the smaller angle between \( c_1 \) and \( c_2 \) is \( \theta \), and that in general for \( i > 2 \) the angle between \( c_i \) and \( c_{i-1} \) is \( \theta \) and the angle between \( c_1 \) and \( c_i \) is \( (i-1)\theta \). Also, \( \theta < \frac{90^\circ}{n} \).
     - Note that the two largest regions will be the same color, and because of our restrictions on \( \theta \), we must have that those two regions cover more than half of the surface area.
     - Thus, the sum of the black and white areas cannot be the same.

\(\blacksquare\)