(4) (50 points) Let $A$ be a $3 \times 9$ grid, with each small cell filled with a positive integer. A sub-grid of $A$ of size $m \times n (1 \leqslant m \leqslant 3, 1 \leqslant n \leqslant 9)$ is called a "good rectangle" if the sum of all its numbers is a multiple of 10. A $1 \times 1$ cell in $A$ is called a "bad cell" if it is not contained in any "good rectangle". Find the maximum number of "bad cells" in $A$.
(4) First, prove that the number of "bad cells" in $A$ is no more than 25.

Use proof by contradiction. Assume the conclusion is not true, then there is at most 1 cell in the grid $A$ that is not a "bad cell". By the symmetry of the grid, we can assume that the first row is all "bad cells".

Let the numbers filled in the $i$-th column from top to bottom in the grid $A$ be $a_{i}, b_{i}, c_{i}, i=1$, $2, \cdots, 9$. Denote
$$
S_{k}=\sum_{i=1}^{k} a_{i}, T_{k}=\sum_{i=1}^{k}\left(b_{i}+c_{i}\right), k=0,1,2, \cdots, 9,
$$

where $S_{0}=T_{0}=0$.
We prove: The three sets of numbers $S_{0}, S_{1}, \cdots, S_{9} ; T_{0}, T_{1}, \cdots, T_{9}$ and $S_{0}+T_{0}$, $S_{1}+T_{1}, \cdots, S_{9}+T_{9}$ are all complete residue systems modulo 10.

In fact, if there exist $m, n, 0 \leqslant m<n \leqslant 9$, such that $S_{m} \equiv S_{n}(\bmod 10)$, then
$$
\sum_{i=m+1}^{n} a_{i}=S_{n}-S_{m} \equiv 0(\bmod 10),
$$

which means the cells from the $(m+1)$-th to the $n$-th column in the first row form a "good rectangle", contradicting the assumption that the first row is all "bad cells".
Similarly, if there exist $m, n, 0 \leqslant m<n \leqslant 9$, such that $T_{m} \equiv T_{n}(\bmod 10)$, then
$$
\sum_{i=m+1}^{n}\left(b_{i}+c_{i}\right)=T_{n}-T_{m} \equiv 0(\bmod 10),
$$

which means the cells from the $(m+1)$-th to the $n$-th column in the second and third rows form a "good rectangle", thus there are at least 2 cells that are not "bad cells", a contradiction.
Similarly, there do not exist $m, n, 0 \leqslant m<n \leqslant 9$, such that
$$
S_{m}+T_{m} \equiv S_{n}+T_{n}(\bmod 10) .
$$

Therefore, the above claim is proven. Hence
$$
\begin{aligned}
\sum_{k=0}^{9} S_{k} & \equiv \sum_{k=0}^{9} T_{k} \equiv \sum_{k=0}^{9}\left(S_{k}+T_{k}\right) \\
& \equiv 0+1+2+\cdots+9 \\
& \equiv 5(\bmod 10),
\end{aligned}
$$

so
$$
\sum_{k=0}^{9}\left(S_{k}+T_{k}\right) \equiv \sum_{k=0}^{9} S_{k}+\sum_{k=0}^{9} T_{k} \equiv 5+5 \equiv 0(\bmod 10),
$$

a contradiction! Therefore, the assumption is false, i.e., the number of "bad cells" cannot exceed 25.
On the other hand, construct the following $3 \times 9$ grid, it can be verified that each cell not filled with 10 is a "bad cell", in this case, there are 25 "bad cells".
\begin{tabular}{|l|l|l|l|l|l|l|l|l|}
\hline 1 & 1 & 1 & 2 & 1 & 1 & 1 & 1 & 10 \\
\hline 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
\hline 1 & 1 & 1 & 10 & 1 & 1 & 1 & 1 & 2 \\
\hline
\end{tabular}

In conclusion, the maximum number of "bad cells" is 25.