Question 5: In an $n \times n$ grid, 101 cells are colored blue. It is known that there is a unique way to cut the grid along the grid lines into some rectangles, such that each rectangle contains exactly one blue cell. Find the minimum possible value of $n$.
(2015, Bulgarian Mathematical Olympiad)
It is easy to give an answer to this problem, but it is not easy to write the process clearly. It should be emphasized that when writing solutions to mathematical problems, especially combinatorial problems, attention should be paid to:
(1) Logical relationships should be clear;
(2) Notations should be declared;
(3) The structure should be clear, and frequently used small conclusions should be stated as lemmas;
(4) Key points should be clearly written (reasonable definitions, clear conclusions).
The minimum possible value of $n$ is 101.
First, prove the more general conclusion below.
Lemma Given an $n \times n$ grid $P$ with $m$ cells colored blue. A "good" partition of the grid $P$ is defined as: if the grid is divided along the grid lines into $m$ rectangles, and each rectangle contains exactly one blue cell. Then the grid $P$ has a unique good partition if and only if the $m$ blue cells form a rectangle.

Proof On the one hand, if the $m$ blue cells form a $m_{1} \times m_{2}\left(m_{1} 、 m_{2} \geqslant 1\right)$ rectangle $A$, cut the four corner cells of $A$ to form rectangles with the four corner cells of the grid $P$; the cells on the four sides of $A$ form $1 \times s(s \geqslant 1)$ rectangles with the cells at the top and bottom (left and right) ends of their respective rows (columns); each cell inside the rectangle $A$ forms a rectangle by itself. This constitutes a good partition $S$ of the grid $P$, and it is easy to see that the grid $P$ has only this unique good partition.

On the other hand, consider the grid $P$ has a unique good partition $S$. Define a grid line as a "partition line" if it can divide the grid $P$ into two rectangles, and each rectangle contains at least one blue cell.

We now prove: every rectangle $A$ containing at least one blue cell has a good partition.

We use mathematical induction on the number of blue cells $k$ in the rectangle $A$ to prove conclusion (1).

If $k=1$, then by the definition of a good partition, conclusion (1) is obviously true.
Assume conclusion (1) holds for $k \leqslant t-1$.
Consider the case $k=t$. In this case, there exists a partition line that divides the rectangle $A$ into two rectangles $A_{1} 、 A_{2}$, and each rectangle contains at least one blue cell. By the induction hypothesis, $A_{1} 、 A_{2}$ each have a good partition. Therefore, the rectangle $A$ has a good partition, and conclusion (1) holds.

By conclusion (1) and the definition of a partition line, every partition line $l$ divides the grid $P$ into two rectangles, each of which has a good partition. Since $S$ is the unique good partition of the grid $P$, the partition $S$ includes all partition lines in the grid $P$.

Let $l_{1}, l_{2}, \cdots, l_{p}$ be all the vertical partition lines from left to right, and $m_{1}, m_{2}, \cdots, m_{q}$ be all the horizontal partition lines from bottom to top. They divide the grid $P$ into some rectangles, and each rectangle contains at most one blue cell. Since $S$ divides the grid $P$ using these partition lines, it results in $m$ rectangles, each containing exactly one blue cell.

Let $l_{0}$ be the vertical grid line closest to $l_{1}$ on the left with no blue cells to its left, and $l_{p+1}$ be the vertical grid line closest to $l_{p}$ on the right with no blue cells to its right. Similarly, define $m_{0} 、 m_{q+1}$ as the horizontal grid lines.
By the above discussion, the distance between $l_{i}$ and $l_{i+1}$
$d\left(l_{i}, l_{i+1}\right)=1(i=0,1, \cdots, p)$;
otherwise, a new partition line can be added between two partition lines.
Similarly, $d\left(m_{j}, m_{j+1}\right)=1(j=0,1, \cdots, q)$.
Therefore, the blue cells form a rectangle with $l_{0} 、 l_{p+1} 、 m_{0} 、 m_{q+1}$ as boundary lines.
The lemma is proved.
By the lemma, since the grid $P$ has a unique good partition, the 101 blue cells must form a rectangle. Therefore, it can only be a $1 \times 101$ rectangle.
Hence, the minimum value of $n$ is 101.