We first claim that the number of “bad cells” in $A$ is no more than $25$. Otherwise, there will be at most one cell in $A$ that is not “bad”. Without loss of generality, we assume the cells in the first row of $A$ are all “bad”. Then let the numbers from top to bottom in the $i$th column be $a_i, b_i, c_i$ ($i = 1, 2, \dots, 9$) in turn, and define
$$S_k = \sum_{i=1}^{k} a_i, \quad T_k = \sum_{i=1}^{k} (b_i + c_i), \quad k = 1, 2, \dots, 9,$$
with $S_0 = T_0 = 0$. We are going to prove that three number groups $S_0, S_1, \dots, S_9$, $T_0, T_1, \dots, T_9$, and $S_0 + T_0, S_1 + T_1, \dots, S_9 + T_9$ each form a complete set of residues modulo $10$:

If there exist $m, n, 0 \le m < n \le 9$ such that $S_m \equiv S_n \pmod{10}$, then
$$\sum_{i=m+1}^{n} a_i = S_n - S_m \equiv 0 \pmod{10},$$
which means that the cells in the first row and from columns $m+1$ to $n$ form a “good rectangle”. But it is a contradiction to the assumption that the cells in the first row are all “bad”.

If there exist $m, n, 0 \le m < n \le 9$ such that $T_m \equiv T_n \pmod{10}$, then
$$\sum_{i=m+1}^{n} (b_i + c_i) = T_n - T_m \equiv 0 \pmod{10}.$$
So the cells ranging from rows $2$ to $3$ and columns $m+1$ to $n$ form a “good rectangle”, which means there are at least two cells that are not “bad”. But it is also a contradiction.

In a similar way, we can also prove that there are no $m, n, 0 \le m < n \le 9$ such that
$$S_m + T_m \equiv S_n + T_n \pmod{10}.$$
Therefore, we have
$$\begin{aligned} \sum_{k=0}^{9} S_k &\equiv \sum_{k=0}^{9} T_k \equiv \sum_{k=0}^{9} (S_k + T_k) \equiv 0 + 1 + 2 + \cdots + 9 \\ &\equiv 5 \pmod{10}. \end{aligned}$$
Then
$$\sum_{k=0}^{9} (S_k + T_k) \equiv \sum_{k=0}^{9} S_k + \sum_{k=0}^{9} T_k \equiv 5 + 5 \equiv 0 \pmod{10}.$$
It is again a contradiction! Therefore, the number of “bad cells” in $A$ is no more than $25$.

On the other hand, we can construct a $3 \times 9$ array in the following and check that each cell in it that does not contain number $10$ is “bad”.

$\boxed{25}$