Find an upper bound for a Linear Recurrence
You say that $N=\lfloor\tfrac{n}{logn}\rfloor$. But since you have a linear recurrence in N, not n, you really want n as a function of N. 

We have $n ≈ N \log n$. 

You substitute this for n and get $n ≈ N \log (N \log n) = N \log N + N \log \log n$. 

$\log \log n$ is small compared to $\log N$, so we ignore it and get $T(N)=T(N-1)+\mathcal{O}(N \log N)$

We get $T(N)$ by summing $\mathcal{O}(k \log k)$ for k = 1 to N, which will be $\mathcal{O}(N^2 \log N)$.