Problem 1121

Lemma: For any broken line $L$ with $A$ and $B$ as its two endpoints, if there are neither horizontal chords of length $a$ nor horizontal chords of length $b$, then there are also no horizontal chords of length $a+b$.

Proof: Suppose there is a chord $PQ$ of length $a$ on the broken line $[A B]$. If we translate the broken line $[A B]$ to the right by a distance of $a$, we obtain the broken line $\left[A_{1} B_{1}\right]$. Then, the two broken lines intersect; otherwise, they do not intersect.

Translate the broken line $[A B]$ to the right by a distance of $a$ to get the broken line $\left[A_{1} B_{1}\right]$, and then translate it to the right by a distance of $b$ to get the broken line $\left[A_{2} B_{2}\right]$. The broken line $\left[A_{1} B_{1}\right]$ will not intersect with the broken line $[A B]$, and the broken line $\left[A_{1} B_{1}\right]$ will not intersect with $\left[A_{2} B_{2}\right]$. Thus, $[A B]$ and $\left[A_{2} B_{2}\right]$ are on opposite sides of the broken line $\left[A_{1} B_{1}\right]$. Therefore, they will not intersect.
Hence, there are no horizontal chords of length $a+b$.
The lemma is proved.

According to the lemma, if the broken line $[A B]$ has no horizontal chords of length $\frac{1}{n}$, then it has no horizontal chords of length $\frac{1}{n} + \frac{1}{n} = \frac{2}{n}$, and no horizontal chords of length $\frac{1}{n} + \frac{2}{n} = \frac{3}{n}$, and so on, no horizontal chords of length $\frac{1}{n} + \frac{n-1}{n} = 1$. Thus, for any positive integer $k$, similarly, there are no horizontal chords of length $k$. Since $[A B]$ has no horizontal chords of any positive integer length, but the segment $A B$ itself is a chord of positive integer length, this is a contradiction.
Therefore, the original problem is proved.