$$\begin{array}{l} L_{1}: y=p_{1} x^{2}+q_{1} x+r_{1}, \\ L_{2}: y=p_{2} x^{2}+q_{2} x+r_{2}, \end{array}$$

where $p_{1}, p_{2} \neq 0$. Suppose the x-coordinates of $A_{i}$ and $B_{i}$ are $a_{i}$ and $b_{i} (i=0,1,2, \cdots, 2 n)$, respectively, then the slope of the line $A_{i} A_{j}$ is
$$\begin{array}{l} \frac{\left(p_{1} a_{j}^{2}+q_{1} a_{j}+r_{1}\right)-\left(p_{1} a_{i}^{2}+q_{1} a_{i}+r_{1}\right)}{a_{j}-a_{i}} \\ =\frac{p_{1}\left(a_{j}^{2}-a_{i}^{2}\right)+q_{1}\left(a_{j}-a_{i}\right)}{a_{j}-a_{i}} \\ =p_{1}\left(a_{j}+a_{i}\right)+q_{1} . \\ (i=0,1,2, \cdots, 2 n-1, j=i+1) \end{array}$$

Similarly, the slope of the line $B_{i} B_{j}$ is
$$p_{2}\left(b_{j}+b_{i}\right)+q_{2}(i=0,1,2, \cdots, 2 n-1, j=i+$$
$1)$.
Thus, from the given conditions, we have
$$\begin{array}{l} p_{1}\left(a_{1}+a_{0}\right)+q_{1}=p_{2}\left(b_{1}+b_{0}\right)+q_{2}, \\ p_{1}\left(a_{2}+a_{1}\right)+q_{1}=p_{2}\left(b_{2}+b_{1}\right)+q_{2}, \\ p_{1}\left(a_{3}+a_{2}\right)+q_{1}=p_{2}\left(b_{3}+b_{1}\right)+q_{2}, \\ p_{1}\left(a_{4}+a_{3}\right)+q_{1}=p_{2}\left(b_{4}+b_{3}\right)+q_{2}, \\ \cdots \ldots . \\ p_{1}\left(a_{2 n-1}+a_{2 n-2}\right)+q_{1}=p_{2}\left(b_{2 n-1}+b_{2 n-2}\right)+q_{2}, \\ \\ p_{1}\left(a_{2 n}+a_{2 n-1}\right)+q_{1}=p_{2}\left(b_{2 n}+b_{2 n-1}\right)+q_{2} . \end{array}$$

Since $A_{0}, B_{0}$ are points on the first parabola $L_{1}$, the slope of $A_{0} B_{0}$ is $p_{1}\left(b_{0}+a_{0}\right)+q_{1}$. At the same time, $A_{0}, B_{0}$ are also points on the second parabola, so the slope of $A_{0} B_{0}$ is $p_{2}\left(b_{0}+a_{0}\right)+q_{2}$. Therefore,
$$p_{1}\left(b_{0}+a_{0}\right)+q_{1}=p_{2}\left(b_{0}+a_{0}\right)+q_{2} .$$

Multiplying (1), (3), ..., (2n-1) by -1 and adding them to (2), (4), ..., (2n) and (*), we get
$$p_{1}\left(b_{0}+a_{2 n}\right)+q_{1}=p_{2}\left(b_{2 n}+a_{0}\right)+q_{2} \text{. }$$

This means $A_{2 n} B_{0} \parallel B_{2 n} A_{0}$.