10.4. In a $100 \times 100$ grid, each cell contains a positive integer. If the sum of the numbers in the cells of a rectangle (composed of several cells) is a multiple of 17, then the rectangle is called "good". We can color the cells of some good rectangles in the grid, with each cell being colored at most once. It is known that for any such number grid, at least $d$ cells can be colored according to the above rule. Find the maximum value of $d$.
10.4. $d_{\max }=9744=100^{2}-16^{2}$.

First, we prove a lemma.
Lemma: In a $1 \times n$ grid, each cell contains a positive integer, then at least $n-16$ cells can be colored. Proof by induction on $n$.
When $n \leqslant 16$, the conclusion is obviously true.
Assume $n=k \geqslant 17$, and for all $n < k$, the conclusion holds.

Suppose the numbers in the leftmost 17 cells are $a_{1}$, $a_{2}, \cdots, a_{17}$. Then among
$$
0, a_{1}, a_{1}+a_{2}, \cdots, a_{1}+a_{2}+\cdots+a_{17}
$$

there exist two numbers such that their difference is divisible by 17, i.e.,
$$
17 \mid\left(a_{i}+a_{i+1}+\cdots+a_{j}\right) .
$$

Thus, the cells from the $i$-th to the $j$-th form a good rectangle A. Removing A from the grid, we get a new $1 \times (k-(j-i+1))$ grid. By the induction hypothesis, the new grid can be expressed as the union of pairwise disjoint good rectangles, except for at most 16 cells. If none of these good rectangles contain both the $(i-1)$-th and $(j+1)$-th cells of the original grid, then they are also good rectangles in the original grid; if one good rectangle contains both the $i$-th and $j$-th cells of the original grid, then its union with A forms a good rectangle in the original grid. In either case, the conclusion holds for $n=k$.

Next, we prove: at least 9744 cells in the grid can be colored.

Consider each column as a large cell, with the number being the sum of the numbers in that column. By the lemma, except for at most 16 columns, the grid can be expressed as the union of several good rectangles of height 100. For the remaining columns, by the lemma, all but at most 16 cells in each column can be colored. Thus, the entire grid has at most $16^{2}=256$ cells that are not colored.

Below is an example where at least 256 cells are not colored.

In a $16 \times 16$ subgrid B of the grid, each cell is filled with 1, and the rest of the cells are filled with 17, then no cell in B can be colored.

In fact, no good rectangle can contain such a cell.

Let A be any good rectangle. If A contains 1, let its intersection with B be an $a \times b$ rectangle, then
$$
a b \equiv 0(\bmod 17),
$$

which contradicts $a, b \leqslant 16$.