In this post, a sudoku like math puzzle is proposed.
The grid must be filled while respecting a unique constraint : the sum of all $3\times3$ sub-squares must equal $2019$.
It is not that difficult to complete the grid, and there are many different solutions. I have noticed empirically that in every solution, the smallest value (among all cells) is at most $4$.
Question : is there an algebraic reason for this ?
The grid is in the picture below.
Note : the statement is easy to prove with a linear solver. Solving the following problem shows that the largest smallest possible value is indeed $4$. $$ \max \left\{ \min{ \{x_{ij}\} } \right\} $$ subject to $$ \sum_{k=i}^{i+2}\sum_{\ell=j}^{j+2} x_{k\ell} = 2019 \quad \forall i,j = 1,...,5 $$ where $x_{ij}$ is the value in cell $(i,j)$.