If the necklace is linear (and not circular), I claim that the answer is 70. An alternating pattern of 7 blue beads and 5 red beads achieves this number, and this clearly satisfies the condition since any segment that contains 8 blue beads will have to contain blue beads from more than one of these sections, and thus contain the 5 red beads in between. To show that this is sufficient, let's just consider the first 99 blue beads. Consider the segments from the 1st to the 8th, the 8th to the 15th, the 15th to the 22nd, and in general the $ 7k+1$st to the $ 7k+8$th bead. There will be 14 of these segments, and each needs to contain 5 red beads. However, a red bead can only lie in one of these segments since they each overlap in exactly one blue bead and no other beads. Thus, 70 red beads are required. For a circular necklace, this requires some fixing, which I'll think about later.

Hamster1800 wrote: If the necklace is linear (and not circular), I claim that the answer is 70. For a circular necklace, this requires some fixing, which I'll think about later. It is circular. First, I believe a necklace must be on somebody's neck (beautiful girls are always preferred), and second, for a string (as I think a linear necklace is properly called) the problem is way too simple.

For each blue bead give it a value of $ 1$ and each red bead give it a value of $ -1$ and for any segment $ s$, $ f(s)$ is the sum of values of these beads. The problem tells us for any segment $ x$ containing $ 13$ beads we know $ f(x) \le 3$. Let there be $ r$ red beads. Now summing over all segment of length $ 13$ we have $ 13(100-r)=%Error. "sumf" is a bad command. (x) \le 3(100+r)$ so $ 14r\ge 1000$ and $ r\ge72$. I am not sure if this bound is the best one though.

Ahwingsecretagent wrote: $ 13(100 - r) = %Error. "sumf" is a bad command. (x) \le 3(100 + r)$ so $ 14r\ge 1000$ and $ r\ge72$. Is not $ 16r \ge 1000$? then $ r \ge 63$

If we, instead of considering segments of 13 beads, consider segments containing exactly 8 blue beads, I believe we can obtain a bound of 72. This is because each of the 100 segments requires 5 red beads, but each red bead will be in 7 segments, so we need at least $ \lceil\frac{500}{7}\rceil = 72$ red beads. It feels like this should be a tight bound (since a 7-5 ratio works for necklaces with $ 7k$ blue beads for some $ k$), but it's not as immediate as it would be if it were 98 blue beads, for example.

Sorry I didn't check my previous post. It should have been $ 13(100-r) = \sum_{x}f(x)\le 3(100+r)$. And yes it should have been $ 16r\ge 1000$ so $ r\ge 63$. Thank you for correcting me mszew. I think this is the correct bound.

Hamster1800 wrote: If we, instead of considering segments of 13 beads, consider segments containing exactly 8 blue beads, I believe we can obtain a bound of 72. This is because each of the 100 segments requires 5 red beads, but each red bead will be in 7 segments, so we need at least $ \lceil\frac {500}{7}\rceil = 72$ red beads. It feels like this should be a tight bound (since a 7-5 ratio works for necklaces with $ 7k$ blue beads for some $ k$), but it's not as immediate as it would be if it were 98 blue beads, for example. One example with $ 72$ is: $ RBRB$ and $ 12$ consecutive blocks of $ BRBRBRBBRBRB$