One. (20 points) In a square with an area of 1, construct a smaller square as follows: divide each side of the unit square into $n$ equal parts, then connect each vertex to the nearest division point of its opposite vertex, as shown in Figure 5. If the area of the smaller square is exactly $\frac{1}{3281}$, find the value of $n$.
I. Draw $C_{1} P \perp A_{1} C$, with the foot of the perpendicular being $P$. Since the area of the square $A B C D$ is 1, we have
$$
A B=B C=C D=1, A_{1} B=\frac{n-1}{n}, C C_{1}=\frac{1}{n} .
$$

By the Pythagorean theorem,
$$
A_{1} C=\sqrt{B C^{2}+A_{1} B^{2}}=\sqrt{1^{2}+\left(\frac{n-1}{n}\right)^{2}} .
$$

It is easy to see that $\mathrm{Rt} \triangle A_{1} B C \backsim \mathrm{Rt} \triangle C P C_{1}$, so
$$
\frac{C_{1} P}{B C}=\frac{C C_{1}}{A_{1} C} \text {. }
$$

Thus, $C_{1} P=\frac{B C \cdot C C_{1}}{A_{1} C}=\frac{1 \times \frac{1}{n}}{\sqrt{1^{2}+\left(\frac{n-1}{n}\right)^{2}}}$
$$
=\frac{1}{\sqrt{2 n^{2}-2 n+1}} \text {. }
$$

Since the area of the small shaded square is $\frac{1}{3281}$, we have
$$
C_{1} P^{2}=\frac{1}{2 n^{2}-2 n+1}=\frac{1}{3281} \text {. }
$$

Simplifying and factoring, we get
$$
(n-41)(n+40)=0 \text {, }
$$

Given $n>0$, we find $n=41$.