Another classic

We claim that if A says "B is a knave", then A and B are different (one is a knight and the other is a knave). We can easily check this case-by-case: if A is a knight, then B is a knave, and if A is a knave, then B shouldn't be a knave, and thus B is a knight.

Thus the statement in the problem is equivalent to the statement that every pair of neighboring people are different. If we number the people $1, 2, 3, \ldots, n$ in order, then we can prove by induction that the odd numbers have the same alignment as 1, and the even numbers have the same alignment as 2; that is, the odd numbers are all knights or all knaves, and so as for the even numbers. (The proof is simple: the base case is proven for 1 and 2, because one has the same alignment as oneself. If we have shown the statement stands for people numbered $k, k+1$ , then $k$ and $k+2$ each has different alignments from $k+1$ , so $k$ and $k+2$ must have the same alignment. If $k$ is odd, then $k+2$ has the same alignment as $k$ , which has the same alignment as 1, so $k+2$ has the same alignment as 1. The proof if $k$ is even is similar.)

Now that we have shown this, we can prove that $n$ is even. Since, if $n$ is odd, then 1 and $n$ neighbor each other and thus should be different, but we proved $n$ and 1 have the same alignment, contradiction. Thus $n$ is even. Moreover, there must be $\frac{n}{2}$ knights and $\frac{n}{2}$ knaves; the odd numbers make $\frac{n}{2}$ of the people, and so as the even numbers, so whichever of them is the knights, we have $\frac{n}{2}$ of them. (Clearly we must have exactly one of the two sets to be knights; if there are zero, then everyone are knaves; if there are two, then everyone are knights, both wrong.)

Thus the conditions are $n$ is even, and $\frac{n}{2} < 100$ . Among the choices, only 192 works.

Given that the sentence is true for everyone, there must be an equal number of knights and knaves, since a knight must have a knave on his left and vice versa. Therefore, if both knight and knaves are less than 100, the total number of participants must be an even number ( $a + a = 2a$ ) and less than 200, so the only solution is $\boxed{192}$