APMO, 2025

Let $(x_{n}, y_{n})$ be the point that Emerald visits after $n$ minutes. Then $(x_{n+1}, y_{n+1}) \in \{(x_{n}+1, y_{n}), (x_{n}, y_{n}+1)\}$. Either way, $x_{n+1}+y_{n+1}=x_{n}+y_{n}+1$, and since $x_{0}+y_{0}=0+0=0$, $x_{n}+y_{n}=n$.

The $n$-th number would be then
$$z_{n}=\left\lfloor x_{n} \alpha+\left(n-x_{n}\right) \beta\right\rfloor \Longrightarrow n \beta+x_{n}(\alpha-\beta)-1<z_{n}<n \beta+x_{n}(\alpha-\beta)$$
in which
$$-2025<x_{n}-y_{n}<2025 \Longleftrightarrow \frac{n-2025}{2}<x_{n}<\frac{n+2025}{2} .$$
Suppose without loss of generality that $\alpha \geq \beta$. Then
$$n \beta+\frac{n-2025}{2}(\alpha-\beta)-1<z_{n}<n \beta+\frac{n+2025}{2}(\alpha-\beta),$$
which reduces to
$$\left|z_{n}-\frac{\alpha+\beta}{2} n\right|<\frac{2025}{2}(\alpha-\beta)+1 .$$
On the other hand, $z_{n+1}=\left\lfloor x_{n+1} \alpha+y_{n+1} \beta\right\rfloor \in\{\left\lfloor x_{n} \alpha+y_{n} \beta+\alpha\right\rfloor,\left\lfloor x_{n} \alpha+y_{n} \beta+\beta\right\rfloor\}$, which implies $z_{n+1} \geq z_{n}$. Since every non-negative integer appears exactly once, in increasing order, it follows that $z_{n}=n$.

Therefore, for all positive integers $n$,
$$\left|n-\frac{\alpha+\beta}{2} n\right|<\frac{2025}{2}(\alpha-\beta)+1,$$
which can only be possible if $\alpha+\beta=2$; otherwise, the left hand side would be unbounded. If $\alpha+\beta=2$, consider $x_{n}=\left\lceil\frac{n}{2}\right\rceil$ and $y_{n}=\left\lfloor\frac{n}{2}\right\rfloor$. If $n$ is even,
$$z_{n}=\left\lfloor\frac{n}{2} \alpha+\frac{n}{2} \beta\right\rfloor=n ;$$
if $n$ is odd,
$$z_{n}=\left\lfloor\frac{n+1}{2} \alpha+\frac{n-1}{2} \beta\right\rfloor=n+\left\lfloor\frac{\alpha-\beta}{2}\right\rfloor,$$
which equals $n$ because $0<\beta \leq \alpha<\alpha+\beta=2 \Longrightarrow 0 \leq \alpha-\beta<2$.