Let $p$ be the smallest prime dividing $n$. Note that then this means $n(p+1) \mid Np$ and $n \mid N$ leading to $n(p+1) \mid N$ from where it follows that $N = kn(p+1)$ for a positive integer $k$. Let $q$ be the smallest prime dividing $N$. The largest number written on the board other than $N$ must be $N/q$, so $N/q = n(1+1/p)$ which implies $q = kp$. Since both $p, q$ are primes we obtain $p = q$ and $k = 1$. Thus $N = n(p+1)$. It's clear that $p = 2$ now, since otherwise $N$ is even and $q = 2 \ne p$, a contradiction. At this point we are done. We claim that all the solutions are $(n, N) = (2^u, 3 \cdot 2^u)$ for positive integers $u$. Suppose that $p$ is the smallest odd prime dividing $n$. Let $i$ be such that $d_i = p$. Note that then $d_{i-1} = 2^v$ for some $v > 0$. Now we know that $\frac{n}{p} + \frac{n}{2^v} \mid 3n$ from where it follows that $p+2^v \mid 3$ which is impossible since the left hand side always exceeds $3$. Hence such a $p$ does not exist and we have $n = 2^u$. It is easy to see that all such $n$ work.