jmc

In order for $N$ to be divisible by 12, $N$ must be divisible by $4$ and $3$. That means the last two digits $AB$ must form a multiple of $4.$ Since $A$ and $B$ are nonzero digits, the smallest possible $14,9AB$ that is divisible by 4 is $14,912.$ Unfortunately, this number is not a multiple of $3$, since $1+4+9+1+2=17.$ However, our next possibility, $14,916,$ is a multiple of $3,$ since $1+4+9+1+6=21.$ Therefore, $\boxed{14,916}$ is our answer.