Design a baseball seam so that all of its 2D surfaces are equally divided? Is it possible to seam a baseball so that the surface area inside the seam will always be equal to the surface area outside of the seam, from all hemispherical viewpoints?
 A: The answer, as the comments under the original question show, depends on the precise formulation of the problem.
Let's regard the baseball as a unit sphere $\Bbb S^2$. We then ask if there is an open subset $X \subset \Bbb S^2$ (the "inside") for all hemispheres $H \subset \Bbb S^2$ we have
$$\mu(X \cap H) = \mu((\Bbb S^2 - X) \cap H) ,$$
where $\mu$ is the standard measure on the sphere. We declare the "seam" to be $\partial X$ and the "outside" to be $Y := \Bbb S^2 - (X \cup \partial X)$, in which case we can write the above condition more symmetrically as $\mu(X \cap H) = \mu(Y \cap H)$.
The key fact is that any $X$ satisfying the condition must be invariant under the antipodal map ${\bf x} \mapsto -{\bf x}$. To see this, first note that we can just as well formulate "equally divided" to mean that signed measure $\chi_X \mu$ satisfies
$$(\chi_X \mu)(H) = \tfrac{1}{2} \mu(H) = \pi$$ is that same for all hemispheres $H$, and a short argument shows that any such measure is invariant under the antipodal map ${\bf x} \mapsto -{\bf x}$, and in our setting that $X$ is invariant under the antipodal map. (Only mentioned in passing in the linked answer is that this can be seen quickly as a satisfying consequence of the Funk transform. This is sufficiently nice that I'd otherwise say more about this here, but this answer will already be long enough.)
If $X$ is invariant under the antipodal map, symmetry gives first
$$(\chi_X \mu)(H) = \tfrac{1}{2} (\chi_X \mu)(\Bbb S^2) = \tfrac{1}{2} \mu(X),$$ so we must have $\mu(X) = 2 \pi$, and then that $X$ has the desired "equally divided" property iff it has these two properties: that (a) $X$ is invariant under the antipodal map and (b) $\mu(X) = 2 \pi$.
There are infinitely many such choices of $X$. We can construct a class of cheap examples, which contains the "lune" example from the comments under the original question: Fix a set $Z$ with measure $\mu(Z) = \pi$ contained inside some hemisphere and set $X := Z \cup -Z$.
On the other hand, for the usual seam on a baseball the "inside" and "outside" are both connected, and we can ask whether this is possible for an "equally divided arrangement" as above (for the examples in the previous paragraph, the inside, $X$, is not connected). The answer is no.
If $X$ is connected, then (since it is open) it is path-connected, so for any choice $x \in X$ there is a path $\gamma$ in $X$ with endpoints $x$ and $-x$, and since $X$ is connected, $-\gamma$ is another path with endpoints $-x$ and $x$. Then the set $C := \gamma \cup -\gamma$ has the property that for any $y \not\in C$, $y$ and $-y$ are in different components of $\Bbb S^2 - C$. Then, since $\Bbb S^2 - C \supset Y$, from taking $y$ in the outside, $Y$, we conclude that $Y$ is not connected.
Finally, we can relax our requirements and ask only for approximate solutions, that is, given $\epsilon > 0$ we can ask for $X$ for which
$$|\mu(X \cap H) - \mu(Y \cap H)| < \epsilon$$ for all $H$. As described in the comments under the original question, this is possible (for all $\epsilon > 0$) even under the condition that $X, Y$ are connected.
Here's a plot of a seam defining an approximate solution satisfying the above condition for some small $\epsilon$:

