The 91st question:
Solution: First, we prove a lemma: In a $2 \times 10$ grid, if $\mathrm{n} \leq 9$ non-adjacent cells are selected, then among the unselected cells, there are more than $\mathrm{n}$ cells adjacent to the selected cells.

Proof of the lemma: Divide the grid into ten $2 \times 1$ sub-grids. Since the selected cells are non-adjacent, in each sub-grid, at most one cell is selected, and if one cell is selected, the other cell is certainly adjacent to it. This means there are already $\mathrm{n}$ cells adjacent to the selected cells. Since $\mathrm{n} \leq 9$, there must be an empty $2 \times 1$ sub-grid.

Find an empty $2 \times 1$ sub-grid adjacent to a selected cell; it must contain a cell adjacent to a selected cell. Therefore, there are at least $\mathrm{n}+1$ empty cells adjacent to the selected cells. The lemma is proved.
Now, let's return to the original problem.
As shown in the figure, we first provide a filling method for $\mathrm{S}=106$: from left to right and from bottom to top, fill the shaded cells with 1, 2, ..., 50, and then from left to right and from bottom to top, fill the blank cells with 100, 99, ..., 51. In this way, the sum of any two adjacent numbers is no greater than $S=106$.
Next, we prove that $S \geq 106$.
Assume there is a filling method such that $S \leq 105$. Divide the large grid into five $2 \times 10$ sub-grids and five $10 \times 2$ sub-grids. Starting from the empty $10 \times 10$ grid, fill in the numbers $100, 99, \ldots, 1$ in descending order. There must be a moment when, after filling in the number $\mathrm{n}_{0}$, either all five $2 \times 10$ sub-grids are non-empty or all five $10 \times 2$ sub-grids are non-empty. Call the first such number $\mathrm{n}_{0}$ the "critical number."

Assume that after filling in the numbers $100, 99, \ldots, 68$ (33 numbers), there are still empty $2 \times 10$ sub-grids and empty $10 \times 2$ sub-grids. Removing these two sub-grids, the remaining 64 cells can be divided into 32 $1 \times 2$ sub-grids. Therefore, among these 33 numbers, there must be two adjacent numbers whose sum is no less than $68+69=137$, which is a contradiction. Thus, the critical number $\mathrm{n}_{0} \geq 68$. And when $\mathrm{n}_{0}$ is filled in, no two numbers are adjacent (otherwise, their sum would be no less than $68+69=137$, which is a contradiction).

When $\mathrm{n}_{0}$ is filled in, assume that all five $2 \times 10$ sub-grids are non-empty. Take any $2 \times 10$ sub-grid; if it contains 10 numbers, then each column is filled with a number. Therefore, before $\mathrm{n}_{0}$ is filled in, all $10 \times 2$ sub-grids already have numbers, which contradicts the fact that $\mathrm{n}_{0}$ is the critical number.

Therefore, when $\mathrm{n}_{0}$ is filled in, each $2 \times 10$ sub-grid contains at most nine numbers. These five $2 \times 10$ sub-grids all meet the conditions of the lemma. At this point, $101-\mathrm{n}_{0}$ numbers have been filled. According to the lemma, these filled numbers have at least $\left(101-n_{0}\right)+5=106-n_{0}$ adjacent empty cells. The largest number in these adjacent empty cells is no less than $106-n_{0}$. The number adjacent to it, which was filled earlier, is no less than $\mathrm{n}_{0}$, and their sum is no less than 106. This contradicts the assumption.
In conclusion, the minimum value of $S$ that satisfies the condition is 106.
The ninety-first question: In a $10 \times 10$ grid, fill in the integers $1, 2, \ldots, 100$, such that the sum of the numbers in any two adjacent cells (cells that share a common edge are called adjacent cells) is no greater than $S$. Find the minimum value of $S$.

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