Why do odd dimensions and even dimensions behave differently?
Solution 1:
For real vector spaces, i.e. $\mathbb{R}^N$, one important difference between odd and even $N$ is that every real polynomial with odd order has a real zero, while there are real polynomials with even order that only have complex zeros.
Thus, a real invertible $N\times N$ matrix always has a non-zero eigenvector if $N$ is odd (because the characteristic polynomial has order $N$, and by the above has a zero). For even $N$, however, there are invertible real matrices with no non-zero eigenvector.
This means that in even-dimensional spaces, there are invertible linear mappings without an invariant one-dimensional subspace, while in odd-dimensional spaces, invertible linear mappings always have a one-dimensional invariant subspace. If you look only at rotations, you get that rotations in an odd-dimensional space always keep at least a line fixed, while in an even-dimensional space they do not.
That's a pretty huge geometric difference between odd- and even-dimensional spaces, that should explain a least some of the items in the question.
Solution 2:
This feels more like a big-list/community-wiki question rather than a bounty problem... Here are two examples:
i) An even-dimensional space supports symplectic forms; an odd-dimensional one does not. [Symplectic = bilinear, alternating, and nondegenerate; bilinear means it corresponds to some $n\times n$ matrix $A$, and then alternating means $A$ is skew-symmetric and nondegenerate means $\det A \neq 0$. For skew-symmetric $A$ we have $\det A = \det (-A^*) = (-1)^n \det(A^*) = (-1)^n \det A$, so for $n$ odd the determinant must vanish; while for even $n$ it's easy to construct nondegenerate examples, such as the block matrix $({\phantom-0 \; I \atop -I \; 0})$ where $I$ is the identity matrix of order $n/2$.]
ii) The Euler characteristic of a sphere of dimension $n$ is $1 + (-1)^n$ which is zero iff $n$ is odd. So you can comb an even-dimensional coconut (whose surface is an odd-dimensional sphere) but not an odd-dimensional one. [To comb the unit ball in ${\bf R}^{2m}$, identify ${\bf R}^{2m}$ with ${\bf C}^m$, and associate to each unit vector $v$ the tangent vector in the direction $iv$, which is orthogonal to $v$.]
[Added later: (i) and (ii) are related because the block matrix $({\phantom-0 \; I \atop -I \; 0})$ is rotation by $i$. But I see that the OP already mentioned the "hairy ball theorem" and linked to the same Wikipedia page. So I offer a third example.]
iii) The real orthogonal groups of even and odd order behave differently as Lie groups: $SO_{2m+1}$ (for $m \geq 1$) is simple of type $B_m$, while $SO_{2m}$ ($m \geq 2$) has nontrivial center $\{\pm 1\}$ and its quotient by the center is simple once $m > 2$ and of type $D_m$.