Does the harmonic series converge if you throw out the terms containing a $9$?

Yes, it's true, as Kempner proved in a short article in 1914; this series, and sometimes variations thereon, are now called Kempner series. A naive argument that groups terms by the number of digits in the denominator gives an upper bound for the limit: $$\sum_{\text{$n$ does not contain $9$}} \frac{1}{n} < 8 \sum_{n = 1}^{\infty} \left(\frac{9}{10}\right)^{n-1} = 80.$$ This estimate is crude, however: The actual value is $22.92067\ldots$. Even computing the value of this series to this accuracy is nontrivial, in part because it converges awfully slowly: Baillie showed that after summing $10^{27}$ terms the remainder still has value $> 1$ (!).

There is nothing special about the number $9$ here, except perhaps that the series omitting it converges slowest among series likewise constructed by omitting the terms of the harmonic series containing a particular digit. One can just as well exclude longer strings of digits, as 5xum mentions in his useful comment; in this general case the series still converges, but (as, in a sense that can be made precise, there are fewer positive integers omitting any given two-digit string than ones omitting $9$) even more slowly than the series omitting $9$.

R. Baillie, Sums of Reciprocals of Integers Missing a Given Digit, The American Mathematical Monthly, Vol. 86, No. 5 (May, 1979), pp. 372-374.

A. J. Kempner, A Curious Convergent Series, The American Mathematical Monthly Vol. 21, No. 2 (Feb., 1914), pp. 48-50.

Just for the removal of nines, you can find the proof here:

https://www.math.wisc.edu/~matz/UgradThesis.pdf

(not my thesis, but it looks like it's proven OK).

It's actually true for any finite string of integers, if I recall correctly. For example, the series converges if we remove all integers which contain the substring $32947902384769234$. The best site I found explaining the phenomenon (first investigated by Kempner, hence the name Kempner series) can be found HERE