Given an $n \times n$ board, with an integer written in each grid, we aim to find the largest $N(n)$ such that for any initial choice of integers, it is possible to make a finite number of moves so that there are at least $N(n)$ even numbers on the board. Each move consists of choosing any grid and adding 1 to all $2n-1$ numbers in its row and column.

The answer is:
$$N(n) = \begin{cases} n^2 - n + 1 & \text{if } n \text{ is odd}, \\ n^2 & \text{if } n \text{ is even}. \end{cases}$$

We will consider everything modulo 2.

### Case 1: $n$ is odd
Firstly, we show that if $n$ is odd, then $N(n) \leq n^2 - n + 1$. Let $r_1, r_2, \ldots, r_n$ and $c_1, c_2, \ldots, c_n$ be the sums of numbers in the rows and columns, respectively. Notice that in each operation, all of these variables change from 0 to 1 or vice versa. Therefore, if the initial configuration is:
$$\begin{pmatrix} 0 & 1 & 1 & \cdots & 1 \\ 0 & 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 0 \end{pmatrix}$$
then we have $(r_1, \ldots, r_n, c_1, \ldots, c_n) = (0, \ldots, 0, 0, 1, \ldots, 1)$. After one operation, this vector becomes $(1, \ldots, 1, 1, 0, \ldots, 0)$, and $(0, \ldots, 0, 0, 1, \ldots, 1)$ after another operation. Hence, there are at least $n-1$ odd numbers every time.

### Case 2: $n$ is even
We will show that the claimed value of $N(n)$ is attainable.

Claim: It is possible to change the parity of one cell and fix all other numbers on the board.

Proof: By symmetry, assume this cell is the top left corner. Notice that by applying the operation to any $n$ cells, none of which lie in the same row or column, the parity of these $n$ cells is changed while all other numbers on the board are fixed. Call this operation $II$. Denote the cell in the $i$-th row and $j$-th column by $(i, j)$. Now, apply operation $II$ to each of the following $n$-tuples:
$$(1,1), (i,1), (1,i), \ldots, (i-1,i-1), (i+1,i+1), \ldots, (n,n), \quad 2 \leq i \leq n.$$
Then all the cells in the first column and first row except the bottom left corner are changed while the other cells are fixed. Apply the given operation to the bottom left corner, and we are done. $\blacksquare$

Now, suppose $n$ is odd. Apply the algorithm for even numbers to the bottom right $(n-1) \times (n-1)$ sub-board. If the first column and first row contain fewer than $n-1$ odd numbers, then we are done. Otherwise, apply an operation to the top left corner, and we are done.

The answer is: $\boxed{\begin{cases} n^2 - n + 1 & \text{if } n \text{ is odd}, \\ n^2 & \text{if } n \text{ is even}. \end{cases}}$.