1. Pavel alternately writes crosses and circles into the cells of a table (starting with a cross). When the table is completely filled, the final score is calculated as the difference $X-O$, where $X$ is the sum of the squares of the counts of crosses in each row and column, and $O$ is the sum of the squares of the counts of circles in each row and column. Determine all possible values of the score achievable for a $67 \times 67$ table.
Consider another solution. Let's think about an $n \times n$ table filled arbitrarily with circles and crosses. We can prove that by changing any circle to a cross, the score increases by $4 n$, and since for a table filled entirely with circles the score is $-2 n^{3}$ and Pavel's filled table contains $\frac{1}{2}\left(n^{2}+1\right)$ crosses, the final score will always be $-2 n^{3} + 4 n \cdot \frac{1}{2}\left(n^{2}+1\right) = 2 n$.

Let the cell being changed be $P$ and assume that the column and row containing $P$ have $s$ and $r$ crosses, respectively. The contribution of this row and column will change from the original

$$
A=r^{2}-(n-r)^{2}+s^{2}-(n-s)^{2}=2 n(r+s)-2 n^{2}
$$

to

$$
B=(r+1)^{2}-(n-r-1)^{2}+(s+1)^{2}-(n-s-1)^{2}=2 n(r+1+s+1)-2 n^{2},
$$

while the contribution of the other rows and columns remains unchanged. Since $B-A=4 n$, we are done.