some context behind the name: the psc (which makes the shortlist) kinda forgot to post this problem onto the psc forum and only realized after the shortlist was determined. After a review we found that the problem was shortlist quality and inserted it between C1 and C2

Solved with MathLuis and YaoAOPS (in particular this is his quick write up).We claim that the minimum nnumber of numbers for a $m,n$ is given by \[f(m,n) = \begin{cases} 2 & \gcd(m,n)>1\\ 3 & \gcd (m,n)=1 \end{cases}\] We first give the construction. If $\gcd(m, n) \ne 1$, two numbers suffice, which is also obviously the minimum possible. First, notice that we can replace the condition with the stronger condition that the sum is constant on any $\gcd(m, n) \times \gcd(m, n)$ square. Then take the numbering which is $0$ on all values unless $x \equiv y \equiv 0 \pmod{\gcd(m, n)}$. Now consider the $\gcd(m, n) = 1$ case.Take a $m \times n$ grid, fill it with $0$s, put $-1$ in two opposite corners and $1$ in the other two opposite corners. uplicate this grid to fill the entire lattice.Then the sum of $1 \times m$ vertical sticks and $1 \times n$ horizontal sticks are both constant, which implies the result because shifting a $m \times m$ by $1$ preserves the sum. Now, we show that when $\gcd(m,n)=1$, it is impossible to cover the board using just 2 numbers. Normalize (shift so that the smaller number is 0 and then scale so that the other is 1) to just being $0, 1$. Note that the sum over a $m \times m$ is $\frac{m^2}{n^2}$ of the sum over a $n \times n$. As such, the sum over a $m \times m$ is either $m^2$ or $0$, giving constancy either way, contradiction.