For each value of $j$, the inner loop runs for $O(N/j)$ iterations. Fixing $i$ and varying over all $j$ in the middle loop, the total number of iterations in the inner loop is
O(N/i + N/(2i) + N/(3i) + N/(ki)) = O(N\log k/i),
where $ki$ is the maximal multiple of $i$ which is at most $N$. We can bound $k$ by $N/i$, and so the expression above equals $O(N\log (N/i)/i) = O(N\log N/i)$. Summing over all $i$ from 1 to $N$, we get $O(N\log^2 N)$.
Using rudimentary calculus (approximating the sum by an integral), you can show that if you sum $N\log (N/i)/i$ over $i$ from 1 to $N$ then you still get $\Theta(N\log^2 N)$, and so the bound above should be tight.