By functoriality there is a map $BG\to B(G/N)$. Let $X$ to be its homotopy fiber. Then you can get a fiber sequence
$\Omega X \to G \to G/N \to X \to BG \to B(G/N)$
using the fact that $\Omega BG = G$ for every (nice enough) topological group $G$. Then, since $N$ is the fiber of $G\to G/N$ we have that $\Omega X = N$. So we only need to show that $X$ is connected to conclude that $X=BN$. Consider then the long exact sequence of homotopy groups induced by $X\to BG\to B(G/N)$. It is
$\pi_1(BG)\to \pi_1(B(G/N))\to \pi_0(X)\to \pi_0(BG)=*$
So we just need to show that the map $\pi_1(BG)\to \pi_1(B(G/N))$ is onto. But this is obvious, since this coincides with $\pi_0(G)\to \pi_0(G/N)$, which is certainly onto.
As a final note, I think that a reference for these matters can be the book by May "Classifying spaces and fibrations".