Solution:

More generally, we will prove that if there are $6n$ beads labeled $n+1, n+2, \ldots, 7n$, there must be three beads in a row whose labels are the side lengths of a triangle. (When $n=289$, this coincides with the problem statement.)

Aiming for a contradiction, assume there are no three beads in a row whose labels are the side lengths of a triangle.

By starting at an arbitrary position on the necklace and counting off three beads at a time, partition the $6n$ beads into $2n$ trios of consecutive beads. Let $S$ be the sum obtained by adding together the smallest two numbers from every trio. (Thus, $S$ is a sum of $4n$ numbers.) Let $T$ be the sum obtained by adding the largest number from every trio.

By our assumption, within each trio, the sum of the two smallest numbers is less than or equal to the largest number. By adding these inequalities across all trios, we see that $S \leq T$.

On the other hand, $S$ can be no smaller than the sum of the $4n$ smallest numbers. Using the formula for the sum of an arithmetic progression, we have
$$S \geq (n+1) + (n+2) + \cdots + (5n) = \frac{(4n)(6n+1)}{2} = 12n^2 + 2n$$
Similarly, $T$ can be no larger than the sum of the $2n$ largest numbers:
$$T \leq (5n+1) + (5n+2) + \cdots + (7n) = \frac{(2n)(12n+1)}{2} = 12n^2 + n$$
Thus $T \leq 12n^2 + n < 12n^2 + 2n \leq S$, which contradicts our earlier claim that $S \leq T$.

We have arrived at a contradiction, so there must in fact be three beads in a row whose labels are the side lengths of a triangle.