Problem 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)
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 $S$. Call a grid line 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. Then 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, $l_{p+1}$ be the vertical grid line closest to $l_{p}$ on the right with no blue cells to its right, and $m_{0} 、 m_{q+1}$ be similarly defined horizontal grid lines.
From 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.