Finite nonabelian groups of odd order For every even $n>4$ there exists nonabelian group. As example of such group we can take dihedral group.
The question is about odd $n$. For some of them there are no nonabelian groups of order $n$ (for example, if $n$ is prime then the group of order $n$ is cyclic and hence abelian).
For what odd $n$ are there known examples of nonabelian finite groups of order $n$?
 A: You might be interested in the result that if n is odd, |G| = n for a finite group G, and if every subgroup of G is normal, then G is abelian. (This does not hold if the hypothesis that n is odd is ommitted as the quaternion group of order 8 demonstrates.) 
A group whose every subgroup is normal is called a Dedekind group. A non-abelian Dedekind group is called a Hamiltonian group. With this terminology the result simply states that a Dedekind group of odd order is abelian. 
The proof is not immediately obvious. It relies on a classification result that states that every Hamiltonian group is a direct product of the quaternion group of order 8, an elemetary abelian 2-group, and a periodic abelian group of odd order. Once this classification result is established, however, the result can be seen easily.
A: It's well-known that for a natural number $n$ with prime factorization
$n=\prod_i p_i^{r_i}$, all groups of order $n$ are abelian if and only if
all $r_i\le 2$ and
$\gcd(n,\Phi(n))=1$ where $\Phi(n)=\prod_i (p_i^{r_i}-1)$.
(See http://groups.google.co.uk/group/sci.math/msg/215efc43ebb659c5?hl=en)
For other $n$ there are non-abelian groups. If some $r_i\ge3$
then we can take a direct product of a non-abelian group of order $p_i^3$ and
a cyclic group. There are always non-abelian groups of order $p^3$;
when $p=2$ take the quaternion group, and when $p$ is odd the group
of upper triangular matrices with unit diagonal over $\mathbb{F}_p$.
Otherwise $G$ will have a factor $pq$ with $p\mid(q-1)$ or $pq^2$ with
$p\mid(q^2-1)$. In the first case the group of all maps $x\mapsto ax+b$
for $a$, $b$, $x\in\mathbb{F}_q$ and $a\ne 0$ has a non-abelian subgroup
of order $pq$. In the second case replace $\mathbb{F}_q$
by $\mathbb{F}_{q^2}$ and then get a non-abelian group of order $pq^2$.
In both cases multiply by a cyclic group to get an order $n$ non-abelian group.
