For the general case it is not a stretch that the above quote in red could also read, "troublemakers can not be a majority of the 2nd members of each pair after eliminating any pairs that have responded "no". Eliminating pairs responding "no" either drops a troublemaker and a truthteller or drops two troublemakers. In both cases this just further promotes the truthtellers as the majority. So with 99 villagers, after 49 questions, consider the worst case where all respond "yes". As Nana7 did, let's collectively call those asking if another is a truthteller the 1st members and those being inquired about the 2nd members. There is obviously a maximum of 49 troublemakers in the 1st members. But in this case, the corresponding members of the second group all must be truthtellers (a clear cut majority which is what we want to prove). Each truthteller swapped with a troublemaker of the 49 1st members puts a corresponding truthteller in the group of second members thus maintaing a majority of truthtellers in that group. When the group of 1st members consists of 24 truthtellers (or more) and 25 troublemakers, the group of 49 2nd members consists of at most 24 troublemakers. So the group of 49 second members always has a majority of truthtellers irregardless of the 99th villager. We can now take the group of 49 second members with an established majority of truthtellers and repeat the process. For less than $100 we can establish a truthteller.
A most excellent puzzle hahaputao!