Competition problem — Combinatorics

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Combinatorics Difficulty 5.3 AIME, harder Prove it Belarus

Each cell of an $n \times n$ table is filled with one of the two signs: «+» and «-». For each $k$ from $1$ to $n$ the amount of pluses in the first $k$ rows is greater than the amount of minuses in the first $k$ columns.
Find the maximal possible number of minuses in the table.

Solution

Answer: $\left[ \frac{n^2 - 1}{2} \right]$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.