so here's what I do know..

if n = 999, then we have 99 0s from the ones digit and 90 0s from the tens digit

if n = 9999, we have 999 0s from the ones digit, 990 0s from the tens digit and 900 0s from the hundreds digit

if

then there are

from the ones digit,

from the tens digit, and so on, thus if

, then we have accumulated

0s, this equals

now, somewhere in between 10^11-1 and 10^12-1, we have more 0s than n, in fact, we have 88,888,888,889 more 0s, so, assuming that the difference is monotone increasing (I believe it is, although I haven't checked formally) it has to happen for some 11 digit number (or it doesn't happen at all)