Suppose we have $n$ balls and $n$ bins, and consider the following process: at stage $k$, we throw $\ln{n}$ balls into the bins, independently at random. We stop after $n/\ln{n}$ stages, when all balls have been thrown.
Let $N_k$ be the set of non-empty bins at the end of stage $k$. I guess that there exists some constant $b$ such that with high probability (that is, with probability that tends to 1 as $n$ grows to infinity; from now on, whp) $|N_k|\ge b\ln{n}$.
My question is as follows: suppose $N_k=\left\{n_k^1,n_k^2,...,n_k^{|N_k|}\right\}$ is ordered in an increasing order by the number of balls, such that if $i<j$, the number of balls in $n_k^j$ is at least the number of balls in $n_k^i$. For $0<c<1$, let $N_k^{(c)}=\left\{n_k^1,n_k^2,...,n_k^{c|N_k|}\right\}$; is it true that for every $0<c<1$ there exists $b=b(c)$ such that whp $\sum_{n\in N_k^{(c)}}|n|\ge bk\ln{n}$? If not, perhaps there exists $b=b(c,k)$ for which it would follow?
If there is such a constant, how fast would $\mathbb{P}\left(|N_k^{(c)}|<bk\ln{n}\right)$ go to $0$?
If $n$ is very large compared to $k$, then $N_k$ will consist of $k\ln(n)$ bins each containing one ball (whp). Thus, $|N_k^{(c)}|$ will be $ck\ln(n)$, and the sum will be $c \binom {k+1}2 \ln(n)$. As $k$ increases, this outpaces $bk \ln(n)$, so $b(c)$ needs only to be chosen to accommodate small values of $k$.
I have not considered the further question (about how fast the probability goes to zero).