In the right-angled coordinate plane $x O y$, $\triangle A_{i} B_{i} A_{i+1}(i=1,2, \cdots)$ are equilateral triangles, and satisfy
$$
\boldsymbol{O A}_{1}=\left(-\frac{1}{4}, 0\right), \boldsymbol{A}_{i} \boldsymbol{A}_{i+1}=(2 i-1,0) \text {. }
$$
(1) Prove that the points $B_{1}, B_{2}, \cdots, B_{n}, \cdots$ lie on the same parabola $\Gamma$, and find the equation of this parabola;
(2) Through the focus $F$ of the parabola $\Gamma$ found in (1), draw two perpendicular chords $A C$ and $B D$. Find the minimum area of the quadrilateral $A B C D$.
Let $B_{n}(x, y)$. Then
$$
\left\{\begin{array}{l}
x=-\frac{1}{4}+1+3+\cdots+(2 n-3)+\frac{2 n-1}{2}=\left(n-\frac{1}{2}\right)^{2}, \\
|y|=\frac{\sqrt{3}}{2}(2 n-1) .
\end{array}\right.
$$

Eliminating $n$ gives $y^{2}=3 x$.
Thus, the points $B_{1}, B_{2}, \cdots, B_{n}, \cdots$ lie on the same parabola $\Gamma: y^{2}=3 x$.
(2) It is easy to see that the focus of the parabola $\Gamma$ is $F\left(\frac{3}{4}, 0\right)$, and $A C$ and $B D$ are not perpendicular to the coordinate axes.
Let the equation of line $A C$ be $x=m y+\frac{3}{4}$.
Substituting into $y^{2}=3 x$ gives $y^{2}-3 m y-\frac{9}{4}=0$.
By Vieta's formulas, we have
$$
\begin{array}{l}
y_{A}+y_{C}=3 m, y_{A} y_{C}=-\frac{9}{4} . \\
\text { Hence }|A C|=\sqrt{1+m^{2}}\left|y_{A}-y_{C}\right| \\
=\sqrt{1+m^{2}} \cdot \sqrt{\left(y_{A}+y_{C}\right)^{2}-4 y_{A} y_{C}} \\
=3\left(1+m^{2}\right) .
\end{array}
$$

Replacing $m$ with $-\frac{1}{m}$ in the above equation gives
$$
|B D|=3\left(1+\frac{1}{m^{2}}\right) \text {. }
$$

Therefore, $S_{\text {quadrilateral } A B C D}=\frac{1}{2}|A C| \cdot|B D|$
$$
\begin{array}{l}
=\frac{9}{2}\left(1+m^{2}\right)\left(1+\frac{1}{m^{2}}\right) \\
\geqslant \frac{9}{2} \times 2 m \times \frac{2}{m}=18 .
\end{array}
$$

The equality holds if and only if $m= \pm 1$. Therefore, the minimum area of quadrilateral $A B C D$ is 18.