We will show that $a_{n}=2 \cdot n!+1$ by induction. Indeed, the claim is obvious for $n=1$, and $(n+1)(2 \cdot n!+1)-n=2 \cdot(n+1)!+1$. Then we wish to find $m \geq 2005$ such that $2(m+1)!\mid 4(m!)^{2}+4 m$ !, or dividing by $2 \cdot m$ !, we want $m+1 \mid 2(m!+1)$. Suppose $m+1$ is composite. Then it has a proper divisor $d>2$, and since $d \mid m$ !, we must have $d \mid 2$, which is impossible. Therefore, $m+1$ must be prime, and if this is the case, then $m+1 \mid m!+1$ by Wilson's Theorem. Therefore, since the smallest prime greater than 2005 is 2011, the smallest possible value of $m$ is 2010.