Team selection test for 20. BMO

Solution:
Let $f(x, y)$ be the number written at the lattice point $(x, y)$. Then
$$f(x, y) = \frac{f(x+1, y) + f(x-1, y) + f(x, y+1) + f(x, y-1)}{4}$$
Assume that not all the numbers are equal. Then there are two points at distance $1$ apart such that the numbers written there are different. Rotating the plane, if necessary, we may assume that $f(x_0+1, y_0) > f(x_0, y_0)$ for some $x_0, y_0 \in \mathbb{Z}$. Set
$$g(x, y) = f(x+1, y) - f(x, y)$$
Then $M = \sup_{x, y \in \mathbb{Z}} g(x, y) \in (0,1]$ and
$$g(x, y) = \frac{g(x+1, y) + g(x-1, y) + g(x, y+1) + g(x, y-1)}{4}$$
In particular, if $g(a, b) \geq M - \varepsilon$, where $\varepsilon > 0$, then
$$\begin{aligned} g(a+1, b) &= 4g(a, b) - g(a-1, b) - g(a, b+1) - g(a, b-1) \\ &\geq 4(M-\varepsilon) - 3M = M - 4\varepsilon \end{aligned}$$
and we get by induction that $g(a+n, b) \geq M - 4^n \varepsilon$ for any $n \in \mathbb{N}$.

Choose now $n \geq \frac{2}{M}$, $\varepsilon \in \left(0, \frac{M}{2 \cdot 4^{n-1}}\right]$ and integers $a$ and $b$ such that $g(a, b) \geq M - \varepsilon$. Then
$$1 > f(a+n, b) > f(a+n, b) - f(a, b) = \sum_{k=0}^{n-1} g(a+k, b) \geq n \frac{M}{2} \geq 1$$
which is a contradiction.