, 2020

Solution:

Clearly $x = y$ works.

Else, WLOG $x < y$, define $d = y - x$, and define $f(z) := ||z - y| - x|$ so our expression reduces to
$$f^{1009}(x) = \left|f^{1009}(0) - y\right|$$
Now note that for $z \in [0, y]$, $f(z)$ can be written as
$$f(z) = \begin{cases} d - z, & 0 \leq z \leq d \\ z - d, & d < z \leq y \end{cases}$$
Hence $f(f(z)) = f(d - z) = z$ for all $z \in [0, d]$. Therefore
$$\left|f^{1009}(0) - y\right| = |f(0) - y| = x$$
If $x > d$ then $f^{1009}(x) < x$ which is impossible (if $f^{1009}(x) \leq d$ then the conclusion trivially holds, and if $f^{1009}(x) > d$ we must have $f^{1009}(x) = x - 1009 d < x$). Therefore $x \leq d$, so $f^{1009}(x) = f(x) = d - x$ and we must have $d - x = x$. Hence $y = 3x$ which is easily seen to work.

To summarize, the possible values of $\frac{x}{y}$ are $\frac{1}{3}, 1, 3$.