Positive real numbers $x$ and $y$ satisfy
$$\Biggl|\biggl|\cdots\Bigl|\bigl||x|-y\bigr|-x\Bigr|\cdots -y\biggr|-x\Biggr|\ =\ \Biggl|\biggl|\cdots\Bigl|\bigl||y|-x\bigr|-y\Bigr|\cdots -x\biggr|-y\Biggr|$$
where there are $2019$ absolute value signs $|\cdot|$ on each side. Determine, with proof, all possible values of $\frac{x}{y}$.

[i]Proposed by Krit Boonsiriseth.[/i]
1. **Initial Setup and Simplification:**
   Given the equation:
   \[
   \Biggl|\biggl|\cdots\Bigl|\bigl||x|-y\bigr|-x\Bigr|\cdots -y\biggr|-x\Biggr| = \Biggl|\biggl|\cdots\Bigl|\bigl||y|-x\bigr|-y\Bigr|\cdots -x\biggr|-y\Biggr|
   \]
   where there are 2019 absolute value signs on each side. We need to determine all possible values of \(\frac{x}{y}\).

2. **Case Analysis:**
   - **Case 1: \(x = y\)**
     If \(x = y\), then both sides of the equation become:
     \[
     \Biggl|\biggl|\cdots\Bigl|\bigl||x|-x\bigr|-x\Bigr|\cdots -x\biggr|-x\Biggr| = \Biggl|\biggl|\cdots\Bigl|\bigl||y|-y\bigr|-y\Bigr|\cdots -x\biggr|-y\Biggr|
     \]
     Simplifying inside the absolute values, we get:
     \[
     \Biggl|\biggl|\cdots\Bigl|\bigl|0\bigr|-x\Bigr|\cdots -x\biggr|-x\Biggr| = \Biggl|\biggl|\cdots\Bigl|\bigl|0\bigr|-y\Bigr|\cdots -x\biggr|-y\Biggr|
     \]
     Since \(x = y\), both sides are equal, and the equation holds true. Therefore, \(\frac{x}{y} = 1\).

3. **Case 2: \(x \neq y\)**
   Without loss of generality, assume \(x > y\). We need to analyze the behavior of the nested absolute values.

   - **Left-Hand Side (LHS):**
     \[
     \Biggl|\biggl|\cdots\Bigl|\bigl||x|-y\bigr|-x\Bigr|\cdots -y\biggr|-x\Biggr|
     \]
     Since \(x > y\), \(|x - y| = x - y\). The innermost absolute value becomes:
     \[
     |x - y| = x - y
     \]
     The next layer:
     \[
     |(x - y) - x| = | - y | = y
     \]
     Continuing this process, we observe that the absolute values alternate between \(x\) and \(y\). Given that there are 2019 absolute value signs, which is an odd number, the final result on the LHS will be:
     \[
     x
     \]

   - **Right-Hand Side (RHS):**
     \[
     \Biggl|\biggl|\cdots\Bigl|\bigl||y|-x\bigr|-y\Bigr|\cdots -x\biggr|-y\Biggr|
     \]
     Similarly, since \(x > y\), \(|y - x| = x - y\). The innermost absolute value becomes:
     \[
     |y - x| = x - y
     \]
     The next layer:
     \[
     |(x - y) - y| = | x - 2y |
     \]
     Continuing this process, we observe that the absolute values alternate between \(x\) and \(y\). Given that there are 2019 absolute value signs, which is an odd number, the final result on the RHS will be:
     \[
     y
     \]

4. **Equating LHS and RHS:**
   From the above analysis, we have:
   \[
   x = y
   \]
   This contradicts our assumption that \(x \neq y\). Therefore, the only possible solution is when \(x = y\).

The final answer is \(\boxed{1}\).