I just need someone to check this argument.
Let $G$ be a nonabelian group of order $2009$. The prime factorization of $2009$ is $7^2 \cdot 41$. Let $n$ be the number of Sylow 7-subgroups.
Then $n \equiv 1$ mod $7$ and $n$ divides $41$. Since $41$ is prime, we must have $n =1$ or $n=41$, but $41 \equiv 6$ mod $7$. Thus $n=1$ and we have a unique Sylow 7-subgroup $H$.
Since $H$ is the unique Sylow 7-subgroup, we have that $H$ is a normal subgroup of $G$. Taking the quotient we have that $|G/H|=41$, so $G/H$ is cyclic of order 41.
Since $G/H$ is cyclic, $G$ is abelian. Thus there is no nonabelian group of order $2009$.