Let's get some values out of the way.
First off, we have to assume the most pessimal option, because that gives us the worst case scenario for who we can save. In that situation the first two logicians to guess are the blind members. The reason this is the worst case is because they cannot set up the rest of the group as they need to.
Now, a single person cannot know the color of their own hat or of the hats behind them. This means that there are three unaccounted for hats by the time we reach the first sighted person. This means there are 97 visible hats in front of the sighted person. Now, this person, unfortunately, has a 50% chance of making the sacrifice play.
The first sighted person looks down the line and counts the red hats. They now know there are N red hats and 97-N blue hats. They then call out which ever of those numbers is higher. Since the numbers are inversely proportional, one will always be higher. From this view, the closest breakdown can be 43 to 44, so our best case in the pessimal option is 44. Now, it's possible that this is the same color as the current logician's hat, but we are going for the pessimal option, so we pretend it isn't. We also pretend that the two blind gents before him also guessed wrong.
Now, this is the pessimal option. This is the value that we are guaranteed to save with this process. But it isn't the most likely option. There is a 2% chance that one of the blind men will end up at the beginning. This means there is a 98% chance the first person in line is sighted. How does this change the field?
Now, there are 99 other people in front of the first person. This means that the minimum number of hats of a given color that he can see in front of him is 49. That is assuming a close to even distribution. If, however, there is a disproportionate number of red or blue, this automatically saves the population represented by the largest number of the same color hat.
Using this method, it is technically possible for all the logicians to be saved, if they all have the same colored hat. This is very, very, very unlikely however.
TL;DR - minimum saved of 44, most likely amount saved around 49
Another alternative would be to slightly bend the rules in our favor.
First, we have to adjust the way the logicians speak. We will assign three choices: no stutter, single stutter, double stutter.
No stutter sounds like
Single stutter sounds like
Double stutter sounds like
No stutter indicates that the hat in front of me is the same color as color I am saying.
Single stutter indicates that the hat in front of me is the opposite color as what I am saying.
Double stutter is assigned to the blind men. This is a way to indicate to the person directly in front of them that they are the blind person, thus the color they are saying has no known connection the hat in front of them.
To illustrate how this would work:
R B B R R R B R B B B R B ...
The first person clearly states
BLUE. They are killed, but the second person also clearly states
BLUE. He is safe. The third person stutters
B-BLUE. She is also safe. The fourth person, hearing the stutter, clearly states
RED. She knows that her hat is the opposite of the given color, so she is safe. The fifth person is blind, so he states
R-R-RED, saving himself and telling the next person that he is blind. The sixth person then clearly states
BLUE. Unfortunately, her hat was red, so she died, but the seventh person can live and carry on the chain.
In our pessimal situation, we can save 97 people using this tactic. This is because the first two people cannot give the next color and the third has to guess. In our pessimal setting, he guesses wrong.
Most likely scenario saves 98 people, as two in front of the blind folks have no way of knowing their actual hat color and are risking their chance of safety.