29.1.3 Lin has a $9 \times 2004$ grid, in which the positive integers from 1 to 2004 are each filled in 9 times. And in each column, the sum of the numbers filled in is no more than 3. Find the minimum possible value of the sum of the numbers in the first row.
Parse $\mathrm{C}_{2003}^{2}+1=2005004$.
Consider a $9 \times n$ grid, in which 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 use induction to prove: the sum of the numbers in the first row is not less than $\mathrm{C}_{n-1}^{2}+1$.

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 $n \geqslant \mathrm{C}_{n-1}^{2}+1$.

Below, we will make the inductive transition. If necessary, we can adjust the order of the columns 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 $S_{i}$ 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 this, we get
$$
\begin{aligned}
S & =\left(n-D_{1}\right)+\left(n-D_{2}\right)+\cdots+\left(n-D_{n}\right) \\
& \geqslant n(n-3)-\left(D_{1}+D_{2}+\cdots+D_{n-3}\right) .
\end{aligned}
$$
(We estimate the sum of $n$ non-negative numbers using the sum of the first $n-3$ terms)
If for any $i \leqslant n-3$, we have $D_{i} \leqslant i+1$, then
$$
S \leqslant n(n-3)-(0+3+4+\cdots+(n-2))=\frac{n^{2}+3 n+4}{2},
$$

which is what we need to prove. Suppose there exists some $k \leqslant n-3$ such that $D_{k} \geqslant k+2$. We point out that in this case, we must have $k \geqslant 2$, so $k+2 \geqslant 4$. Since there are at least $k+2$ numbers in the first row that are less than $k$, all 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 $n-1 \geqslant k+2 \geqslant 4$, the sum of the numbers in the first row of the first part is not less than $\mathrm{C}_{k+1}^{2}+1$. 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 $(k+2)(n-k-2)+\mathrm{C}_{n-k-3}^{2}+1$. Adding these two estimates, we get
$$
S \geqslant \mathrm{C}_{k+1}^{2}+1+(k+2)(n-k-2)+\mathrm{C}_{n-k-3}^{2}+1=\mathrm{C}_{n-1}^{2}+3,
$$

which is even stronger than what we need to prove.
We also provide an example that can achieve the minimum possible value:
\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 \\
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
\end{tabular}