Let $n\geq 3$ be an odd number. We will play a game using a $n$ by $n$ grid. The game is comprised of $n^2$ turns, in every turn, we will perform the following operation sequentially.

$\bullet$ We will choose a square with an unwritten integer, and write down an integer among 1 through $n^2$. We can write down any integer only at once through the game.

$\bullet$ For each row, colum including the square, if the sum of integers is a multiple of $n$, then we will get 1 point (both of each sum is a multiple of $n$, we will get 2 points).

Determine the maximum possible value of the points as the total sum that we can obtain by the end of the game.
1. **Understanding the Problem:**
   We are given an \( n \times n \) grid where \( n \) is an odd number (\( n \geq 3 \)). We need to fill this grid with integers from 1 to \( n^2 \) such that each integer is used exactly once. The goal is to maximize the points obtained by making the sum of integers in each row and column a multiple of \( n \). Each row or column that sums to a multiple of \( n \) gives 1 point.

2. **Modular Arithmetic Insight:**
   Consider the numbers modulo \( n \). Since \( n \) is odd, each residue class modulo \( n \) will appear exactly \( n \) times among the numbers \( 1, 2, \ldots, n^2 \). Ignoring the numbers divisible by \( n \), we need to determine the maximum number of points a row or column can contribute.

3. **Maximum Points Calculation:**
   - Assume the current sum of a row or column is \( 0 \mod n \). To reach \( 0 \mod n \) again, we need to place at least 2 numbers.
   - The maximum number of points obtainable without considering the numbers divisible by \( n \) is:
     \[
     \frac{n-1}{2} \times 2n = n(n-1)
     \]
   - Each number divisible by \( n \) can contribute at most 2 points (one for the row and one for the column). Since there are \( n \) such numbers, the additional points are:
     \[
     2n
     \]
   - Therefore, the total maximum points are:
     \[
     n(n-1) + 2n = n(n+1)
     \]

4. **Constructing the Grid:**
   - Arrange the numbers in the grid such that the square \((a, b)\) has the number \( (a + b) \mod n \) in it.
   - Select any non-zero residue. Choose an "opposite" residue in the same row or column until you return to the original field. This contributes 0 points for the first move, 1 point for each subsequent move, and 2 points for the last move, totaling \( 2n \) points and returning every row and column to \( 0 \mod n \).
   - Repeat this for every pair of non-zero residues and finally for the zeros.

5. **Verification:**
   - For each pair of non-zero residues, the total points are:
     \[
     2n \times \frac{n-1}{2} + 2n = n(n+1)
     \]

Thus, the maximum possible value of the points is \( n(n+1) \).

The final answer is \( \boxed{ n(n+1) } \)