6. 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.
6. Solution: The minimum possible value is $C_{2003}^{2}+1=2005004$.

Consider a $9 \times n$ grid, where the positive 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 obviously true. 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 adjust the order of the numbers so that the numbers in the first row are in non-decreasing order. Therefore, we can assume that the numbers in the first row are already in non-decreasing order. Let $S_{i}$ be 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 $S=S_{1}+S_{2}+\cdots+S_{n}$.

Rewriting the above expression, 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 $S \geqslant n(n-3)-[0+3+4+\cdots+(n-2)] = \frac{n^{2}-3n+4}{2}$, which is what we need to prove.

Assume there exists some $k \leqslant n-3$ such that $D_{k} \geqslant k+2$, in which case $k \geqslant 2$. 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 subsequent columns are 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 $n-1 \geqslant k+2 \geqslant 4$, 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 $S \geqslant C_{k+1}^{2}+1+(k+2)(n-k-2)+C_{n-k-3}^{2}+1 = C_{n-1}^{2}+3$.
Below is an example that achieves the minimum possible value (see the table below):
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline 1 & 1 & 1 & 2 & 3 & 4 & $\cdots$ & $k$ & $\cdots$ & 1998 & 1999 & 2000 & 2001 & 2001 \\
\hline 1 & 2 & 3 & 4 & 5 & 6 & $\cdots$ & $k+2$ & $\cdots$ & 2000 & 2001 & 2002 & 2003 & 2004 \\
\hline 1 & 2 & 3 & 4 & 5 & 6 & $\cdots$ & $k+2$ & $\cdots$ & 2000 & 2001 & 2002 & 2003 & 2004 \\
\hline 1 & 2 & 3 & 4 & 5 & 6 & $\cdots$ & $k+2$ & $\cdots$ & 2000 & 2001 & 2002 & 2003 & 2004 \\
\hline 1 & 2 & 3 & 4 & 5 & 6 & $\cdots$ & $k+2$ & $\cdots$ & 2000 & 2001 & 2002 & 2003 & 2004 \\
\hline 1 & 2 & 3 & 4 & 5 & 6 & $\cdots$ & $k+2$ & $\cdots$ & 2000 & 2001 & 2002 & 2003 & 2004 \\
\hline 1 & 2 & 3 & 4 & 5 & 6 & $\cdots$ & $k+2$ & $\cdots$ & 2000 & 2002 & 2002 & 2003 & 2004 \\
\hline 2 & 3 & 4 & 5 & 6 & 7 & $\cdots$ & $k+3$ & $\cdots$ & 2001 & 2002 & 2003 & 2003 & 2004 \\
\hline 2 & 3 & 4 & 5 & 6 & 7 & $\cdots$ & $k+3$ & $\cdots$ & 2001 & 2002 & 2003 & 2004 & 2004 \\
\hline
\end{tabular}