Let $s$ be an integer greater than 1. For each natural number $n$, let $\omega(n)$ be the number of prime divisors (counted with multiplicities) of $n$ modulo $s$. For $i \in \{0,1,\dots, s-1\}$ and a positive integer $k$, set 
$$c\_i(k) = |\{x \in \{1, 2, \dots, k\} : \omega(x) = i\}|.$$
Is it true that $\frac{c\_i(k)}{k} \rightarrow \frac{1}{s}$ as $k \rightarrow \infty$ for all $i \in \{0,1, \dots, s-1\}$?



proportion of positive integers which number of prime divisors has a special remainder.
Yes, this is true.


More precisely, it is known that, for fixed $s$ and $k \to \infty$, one has 
`$$ \frac{c_i(k)}{k} = \frac{1}{s} + O(\frac{1}{ (\log k)^A}) $$`
for some $A>0$ .


See 'On the residue class distribution of the number of prime divisors of an integer' by Coons and Dahmen; preprint at <http://arxiv.org/abs/0906.1029> 


They also discuss the nature of the error-term in more detail, and in particular show that for $s>2$ it cannot be `$ O(1/ k^B) $` for $B>0$.


Unrelated side remark: since you define $\omega$ precisely this is not a big deal, but using the standard notation for the number of distinct prime divisors for something else is somewhat 'dangerous' (at least, it almost lead me to give a wrong answer).


