Make the additional assumptions that both $n$ and $k \to +\infty$ (as the common variable goes to some limit). Let $g(x) = x/\log x$. Note that this is increasing for $x \ge e$ and $g(t \log t) = \frac{t \log t}{\log t + \log \log t} < t $ for $t > e$, and $g(t \log t) = t + o(t)$ as $t \to \infty$.
1) Suppose $n = o(k \log k)$. Given $K > 0$, take $\delta = 1/(2 K)$.
For $k$ and $n$ sufficiently large we have $n < \delta k \log k$ so $ k > g(k \log k) > g(n/\delta) = \frac{n}{\delta \log(n/\delta)} > \frac{K n}{\log n}$ Therefore $k = \omega(n/\log n)$.
2) Conversely, suppose $k = \omega(n/\log n)$. Given $\epsilon > 0$, take $B = 2/\epsilon$. For $k$ and $n$ sufficiently large we have $k > 2 B g(n)$. Now $g(k/B \log (k/B)) = k/B + o(k/B) > g(n)$ , so $n < k/B \log(k/B) < \epsilon k \log k$, for sufficiently large $k$ and $n$. Thus $n = o(k \log k)$