Smallest polyhedron with an odd number of all n-gon faces So this is a question I actually "know" the answer to (in the sense that I know broadly what the answer is, but I'm missing crucial details and expertise to find an exact unique answer).
A long while ago, someone posed this question to me, and I found that the smallest example of a polyhedron with an odd number of faces where each face is an n-gon is an enneahedron with 4-gon faces.  I don't remember where I found this particular assertion but I suspect it might have been on OEIS or something.
At the time I had hunted down a page that had a full explanation and image of this object, but unfortunately it is now a 404.
The motivation for this question is I recently got into a discussion about different rhombuses that appear in various geometrical shapes, and I thought I remembered the 4-gons in this object being rhombuses (however, I might be completely mistaken), and I was wondering after their ratio if they were.
 A: Let $e_i$ be the counts of edge type $i$, $f_i$ the counts of face type $i$ and $n_i$ number of edges of the $i$-th face type. Then you will have
$$2\sum e_i=\sum n_i\,f_i$$
Thus, when asking for a single face type (i.e. at the right hand side the sum is breaking down), your quest cannot be solved for any odd $n$, because then $f$ is being forced to be even.
So indeed, the smallest possible value of $n$ within your quest would be 4.
--- rk
A: A partial answer:
To prove that $9$ is the fewest odd number of faces a polyhedron with all faces having the same number of sides we can proceed as follows:
Start with Euler's formula $V+F=E+2$.  If there are $k$ faces of $n$ sides each this becomes $V+k=\frac {nk}2+2$ because each edge is counted on two faces.  If $k$ is odd $n$ must be even to make the fraction integral.  
The total of all the angles at all the vertices must be $4\pi$ less than $2\pi V$ to make it close.  If the faces are $4$ sided we must have two less faces than vertices.  As you must have at least three faces meet at each vertex, each vertex can only contribute a maximum average of $\frac \pi 2$ angle deficit, so there must be at least $9$ vertices.  Finally one has to show that there is no solution with $9$ vertices and $7\ 4$ sided faces.  I don't have an approach for that part.
