Question 28: An $n \times n$ grid table is composed of $n^{2}$ cells, where $n$ is a positive integer. All edges of each cell are drawn, and the diagonals of some cells are also drawn, with at most one diagonal being drawn in any cell, and the diagonals (if any) of any two adjacent cells with a common edge are in different directions. Find all values of $n$ for which there exists a drawing method that allows one to start from the bottom-left vertex and draw the entire figure in one stroke (i.e., each edge or diagonal is traversed exactly once).

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Question 28:
Answer: When $\mathrm{n}=1, 2, 3$, the following three figures satisfy the conditions:

Below is the proof: When $\mathrm{n} \geq 4$, there does not exist a drawing method that satisfies the conditions.
If the conclusion is not true, then there are at most two odd vertices in the drawn graph. According to the pigeonhole principle, there must exist a side of the grid with $\mathrm{n}-1$ points (excluding the four vertices of the grid) all being even vertices. Without loss of generality, assume that the $\mathrm{n}-1$ points on the left side are all even vertices, as shown in the figure below. Thus, from $A_{1}, A_{2}, A_{3}$, each point must draw exactly one diagonal.

If the diagonal drawn from $\mathrm{A}_{1}$ points to the upper right, then the diagonal drawn from $\mathrm{A}_{2}$ must point to the lower right, which means $\mathrm{A}_{3}$ cannot draw a diagonal, leading to a contradiction!
If the diagonal drawn from $\mathrm{A}_{1}$ points to the lower right, then $\mathrm{A}_{2}$ cannot draw a diagonal, still leading to a contradiction!
In summary, the required $n=1,2,3$.