8. (30th Russian Mathematical Olympiad) There is a $9 \times 2004$ grid, in which the positive integers from 1 to 2004 are each filled in 9 times, and the difference between the numbers in each column does not exceed 3. Find the minimum possible value of the sum of the numbers in the first row.
8. $C_{2003}^{2}+1=2005004$

Consider a $9 \times n$ grid, where the integers from 1 to $n$ are each filled in 9 times, and the difference between the numbers in each column does not exceed 3. We will prove by mathematical induction that the sum of the numbers in the first row is not less than $C_{n-1}^{2}+1$.
When $n \leqslant 4$, the conclusion is obvious, because each number in the grid is not less than 1, and when $n \leqslant 4$, we have $n \geqslant C_{n-1}^{2}+1$.
Next, we perform the inductive step. If necessary, we can rearrange the columns so that the numbers in the first row are in non-decreasing order, so we can assume that the numbers in the first row are already in non-decreasing order. Let $S_{i}$ denote the number of numbers in the first row that are not less than $i$, and let $D_{i}=n-S_{i}$. Thus, $S_{1}=n, D_{1}=0$, and the sum of the numbers in the first row is
$$
S=S_{1}+S_{2}+\cdots+S_{n}.
$$

Rewriting the above, we get
$$
S=\left(n-D_{1}\right)+\left(n-D_{2}\right)+\cdots+\left(n-D_{n}\right) \geqslant n(n-3)-\left(D_{1}+D_{2}+\cdots+D_{n-3}\right).
$$

If for any $i \leqslant n-3$, we have $D_{i} \leqslant i+1$, then $S \geqslant n(n-3)-[0+3+4+\cdots+(n-2)]=\frac{n^{2}-3 n+4}{2}$. This is what we need to prove.
Assume there exists some $k \leqslant n-3$ such that $D_{k} \geqslant k+2$. In this case, $k \geqslant 2$, so $k+2 \geqslant 4$. Since there are at least $k+2$ numbers in the first row that are less than $k$, the numbers in the first $k+2$ columns do not exceed $k+2$. This means that all such numbers are in the first $k+2$ columns, so the numbers in the remaining columns are not less than $k+3$. Now, we divide the entire grid into two parts: the first part consists of the first $k+2$ columns, and the rest form the second part. Since $n-1 k+2 \geqslant 4$, the sum of the numbers in the first row of the first part is not less than $C_{k+1}^{2}+1$. If we subtract $k+2$ from each number in the second part, we get a grid with $n-(k+2) \geqslant 1$ columns that satisfies the problem's conditions. By the inductive hypothesis, the sum of the numbers in the first row of this part is not less than $(k+2)(n-k-2)+C_{n-k-3}^{2}+1$.
Adding these two estimates, we get
$$
S \geqslant C_{k+1}^{2}+1+(k+2)(n-k-2)+C_{n-k-3}^{2}+1=C_{n-1}^{2}+3.
$$

We also provide an example that can achieve the minimum possible value. As shown in the table below: