jmc

The $231$ cubes which are not visible must lie below exactly one layer of cubes. Thus, they form a rectangular solid which is one unit shorter in each dimension. If the original block has dimensions $l\times m\times n$, we must have $(l-1)\times (m-1)\times (n-1)=231$. The prime factorization of $231=3\cdot 7\cdot 11$, so we have a variety of possibilities; for instance, $l-1=1$ and $m-1=11$ and $n-1=3\cdot 7$, among others. However, it should be fairly clear that the way to minimize $l\cdot m\cdot n$ is to make $l$ and $m$ and $n$ as close together as possible, which occurs when the smaller block is $3\times 7\times 11$. Then the extra layer makes the entire block $4\times 8\times 12$, and $N=\boxed{384}$.