Question 4. No figure, divide each side of a single square into $\mathrm{n}$ equal parts, then connect each vertex to the nearest division point of the opposite vertex, thus creating a smaller square inside the square. If the area of the smaller square (shaded in the figure) is exactly $\frac{1}{1985}$, find the value of $\mathrm{n}$.
Solve as shown in the figure. $\mathrm{DM}=\frac{\mathrm{n}-1}{\mathrm{n}}-$
$$
\begin{array}{l}
\therefore A M= \\
\sqrt{1^{2}+\left(\frac{n-1}{n}\right)^{2}} \\
=\frac{1}{\cos \angle D E H=\cos \angle D A M=\frac{D A}{A M}} \\
\sqrt{1+\left(1-\frac{1}{n}\right)^{2}}
\end{array}
$$

The area of the small square in the middle $\mathrm{S}=\mathrm{GF}^{2}=\mathrm{EH}^{2}=(\mathrm{DE} \cdot \cos \angle \mathrm{DEH})^{2}$
$$
=\frac{1}{1+\left(1-\frac{1}{n}\right)^{2}} \cdot \frac{1}{n^{2}}=\frac{1}{1985},
$$

which gives $2 n^{2}-2 n+1=1985$. Solving this, we get $n_{1}=32$, $\mathrm{n}_{2}=-31$ (discard).
Therefore, $\mathrm{n}=32$.