Grrr, commenting requires 50 rep. So this answer is actually a comment on chris's answer.
Since the questioner probably doesn't care about all the directories, only the worst ones, then using sort is likely very expensive overkill.
find . -type d |
echo "$(ls "$line" | wc -l) $line"
perl -a -ne'next unless $F>=$max; print; $max=$F' | less
This isn't as complete as your version, but what this does is print lines if they're larger than the previous maximum, greatly reducing the amount of noise printed out, and saving the expense of the sort.
The downside of this is if you have 2 very large directories, and the first happens to have 1 more inode than the 2nd, you'll never see the 2nd.
A more complete solution would be to write a smarter perl script that keeps track of the top 10 values seen, and prints those out at the end. But that's too long for a quick serverfault answer.
Also, some midly smarter perl scripting would let you skip the while loop - on most platforms, ls sorts the results, and that can also be very expensive for large directories. The ls sort is not necessary here, since all we care about is the count.