2. Given an odd number $n \geqslant 3$. A game is played on an $n \times n$ grid. The game consists of $n^{2}$ steps, each step following these rules:
(1) Choose an empty cell and fill it with a positive integer that is no less than 1 and no more than $n^{2}$. Each positive integer can only be used once throughout the process;
(2) In the grid, if the sum of all numbers in the row or column where the filled number is located is a multiple of $n$, then 1 point is scored. If the sum of all numbers in both the row and column where the filled number is located are multiples of $n$, then 2 points are scored.
Find the highest score that can be achieved by completing this game.
2. $n(n+1)$.

Obviously, whether the filled number can score depends only on the remainder obtained by dividing it by $n$. Therefore, each positive integer $k(0 \leqslant k \leqslant n-1)$ can be used exactly $n$ times.
Next, we prove that the highest score obtained is $n(n+1)$.
First, construct a method to achieve a score of $n(n+1)$.
Fill in the remainder of $i+j$ divided by $n$ at the position of the $i$-th row and $j$-th column. Fill in the numbers in the following order: 0 is filled in $n$ times, 1 is filled in $n$ times, $n-1$ is filled in $n$ times, 2 is filled in $n$ times, $n-2$ is filled in $n$ times, $\cdots \cdots$ This way, when 0, $n-1, n-2, \cdots, \frac{n+1}{2}$ are filled in, each time 2 points are scored, thus the final score is $n(n+1)$.
Next, we prove that the highest score obtained does not exceed $n(n+1)$. Let $A_{0}=B_{0}=0$, for any positive integer $i(1 \leqslant i \leqslant n^{2})$, if the number filled in during the $i$-th operation makes the sum of the numbers in this row a multiple of $n$, then $A_{i}=A_{i-1}+1$; otherwise, $A_{i}=A_{i-1}$.
Similarly, for columns, define $B_{i}\left(1 \leqslant i \leqslant n^{2}\right)$.
Therefore, the final score is $A_{n^{2}}+B_{n^{2}}$.
Next, consider the maximum value that $A_{n^{2}}$ can take.
For each $i\left(0 \leqslant i \leqslant n^{2}\right)$, let $C_{i}$ represent the number of rows where the sum of the numbers filled in is not a multiple of $n$ after the $i$-th operation. Let $C_{0}=0$, then, it can be seen that when 0 is filled in during the $i$-th operation, $A_{i}+\frac{1}{2} C_{i}$ increases by at most 1 compared to $A_{i-1}+\frac{1}{2} C_{i-1}$, otherwise, it increases by at most $\frac{1}{2}$.
Since 0 is filled in $n$ times, and non-zero numbers are filled in $n^{2}-n$ times, therefore,
$A_{n^{2}}+\frac{1}{2} C_{n^{2}} \leqslant n+\frac{n(n-1)}{2}=\frac{n(n+1)}{2}$.
Since $C_{n^{2}}$ is a non-negative integer, therefore,
$A_{n^{2}} \leqslant \frac{n(n+1)}{2}$.
Similarly, $B_{n^{2}} \leqslant \frac{n(n+1)}{2}$.
Thus, $A_{n^{2}}+B_{n^{2}} \leqslant n(n+1)$.
Therefore, the maximum score is $n(n+1)$.