Let $n$ and $s$ be the numbers of natives and strangers, respectively.
We are told that $n > s$ and that natives always tell the truth but strangers lie when they wish.
As stated, the puzzle requires an inquirer to ask just one individual ("ask anyone") a series of questions to separate the set of all natives from the set of all strangers.
Now, had the strangers been incapable of telling the truth, you would be able to craft questions that produce consistent answers (e.g. "How many natives would a liar never say there are?"). But since the strangers can tell the truth but also can lie, their answers are unreliable. So if the inquirer picked a stranger, the answers need not take any element of the inquiry into account. In that case, no set of questions would give you certainty and the puzzle has no answer. At best, the inquirer can tell they are talking to a stranger, but cannot compel the stranger to produce the correct answer.
However, consider a variant where the inquirer can ask multiple people multiple questions. Then one strategy is to ask each person, "Who are the natives?" and go with the majority. Here, it doesn't matter what the strangers say - if they tell the truth, they contribute to the majority, and if they lie, they are in the minority anyway. For this strategy to work, the number of natives questioned must be greater than the number of strangers questioned, so in the worst case, $2s + 1$. Since $n>s$, we also have $n+s > s+s \geq 2s+1$, so there are enough people to ask. However, this assumes knowledge of $s$, which may not be the case.
If $n+s$ is odd, then from $n>s$, we have $n+s \geq (s+1) + s$, i.e. $n+s \geq 2s+1$.
If $n+s$ is even, then from $n>s$, we have $n+s \geq (s+2) + s$, i.e. $n+s \geq 2s+2$, or $n+s-1 \geq 2s+1$.
So according to this strategy, the answer is $2s+1$ questions if you know $s$, otherwise it is $n+s$ for an odd population and $n+s-1$ for an even population.
Is there a better strategy?
Since strangers provide no information, any optimal strategy will need to include asking at least one native and be able to tell when at least one native has answered. In the worst case, an optimal strategy will require $s+1$ questions, but since the inquirer doesn't know $s$, we're back to a total of $n+s$ questions for odd populations and $n+s-1$ for even populations.
