Answer: By even $k$ the first player collects $(3k + 2)$, the second $k$; by the odd $k$ the first $(3k + 1)$, second $(k + 1)$.

Let $n = 2k$.

In the beginning we can look into such a strategy for every player. The first player fills numbers in arbitrary way. In this way he collects $k$ points. The second -- in this way to last line collects $(k-1)$ points, but in the last line he fills numbers to win in every column. In this way he collects $2k + (k-1) = 3k-1$ points. Only we have to find out who with this strategy will win in last line. As product of all numbers in table is "+1" because product in all columns is $1$ and there is even quantity. Odd lines have product "+1"; in this way the product of all "even" lines is "+1". There are $k$ pieces.

In this way with even $k$ the product of last line will be $1$ too. Definitely first player collects at least $k$ points, and second -- $3k$. In attempt of changing strategy, each of them can only reduce his result, because the opponent can use this for improvement of his result. In the way of odd $k$ the product of last line must be "+1", i.e. the first player will win. The final result in this way will be: first -- $(k+1)$, second -- $(3k-1)$. Whether the second can collect more? In way of another strategy, first player will collect his $k$ points for the lines. If the second loses only one column or line, he would collect no more than $(3k-1)$. But he can't win all of them, in consequence of valuation mentioned here. The first can't collect more than $(k+1)$, because the second can collect $(3k-1)$ at least.

Let $n = 2k + 1$.

The strategy remains here, but only in result of odd size of the table, the last move does the first player. In this way he collects his points for all $(2k+1)$ columns and $k$ lines, without last. The second wins his $k$ columns. We only have to check who will win in the last line with these strategies. The product of columns of all numbers is "+1" -- all of them positive. The product of all lines, except last is $(-1)^k$. That's why with even $k$ the first wins the last line too, but with odd -- the second wins. Thereby answer is: in the way of paired $k$ the first collects $(3k+2)$, the second -- $k$; with odd $k$, the first -- $(3k+1)$, the second -- $(k+1)$.