67th Czech and Slovak Mathematical Olympiad

Let $n=67$ and denote by $k=\frac{1}{2}(n^2+1)$ the total number of crosses in the table. A row containing $a$ crosses and $n-a$ circles contributes $a^2-(n-a{)}^2=2n\cdot a-n^2$ to the total score and thus all the $n$ rows combined contribute $$2n\cdot k-n\cdot n^2=2n\cdot \frac{n^2+1}{2}-n^3=n$$ to the total score. Likewise, columns contribute $n$. Hence the total score is always equal to $2n=134$.

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Alternative solution.

Consider an $n\times n$ table filled with arbitrarily many crosses and circles. We show that replacing any circle by a cross increases the score by $4n$. Since the score for a table filled with all circles equals $-2n^3$ and Paul's table contains $\frac{1}{2}(n^2+1)$ crosses, the final score will always be equal to $-2n^3+4n\cdot \frac{1}{2}(n^2+1)=2n$.

Consider any cell containing a circle and denote by $r$ and $c$ the number of crosses in its row and column, respectively. The contribution of this row and column changes from $$A=r^2-(n-r{)}^2+s^2-(n-s{)}^2=2n(r+s)-2n^2$$ to $$B=(r+1{)}^2-(n-r-1{)}^2+(s+1{)}^2-(n-s-1{)}^2=2n(r+1+s+1)-2n^2$$ and the contribution of other rows and columns doesn't change. Since $B-A=4n$, we are done.