Three. (50 points) Given that the difference between any two adjacent interior angles of a convex $n$-sided polygon is $20^{\circ}$. Find the maximum value of $n$.


(1) Let the convex $n$-sided polygon be denoted as $A_{1} A_{2} \cdots A_{n}$. Then,
$$
\begin{array}{l}
A_{2}=A_{1} \pm 20^{\circ}, \\
A_{3}=A_{2} \pm 20^{\circ}=A_{1} \pm 20^{\circ} \pm 20^{\circ}, \\
\cdots \cdots \\
A_{k}=A_{1} \pm 20^{\circ} \pm 20^{\circ} \pm \cdots \pm 20^{\circ} \\
=A_{1}+20^{\circ} p_{k}-20^{\circ} q_{k},
\end{array}
$$

where $p_{k}$ is the number of “+” signs, and $q_{k}$ is the number of “-” signs, with $p_{k}+q_{k}=k-1$.
In the above equation, let $k=n+1$, i.e.,
$$
\begin{array}{l}
A_{n+1}=A_{1}+20^{\circ} p_{n+1}-20^{\circ} q_{n+1} \\
=A_{1}+20^{\circ} p_{n+1}-20^{\circ}\left(n-p_{n+1}\right)=A_{1} \\
\Rightarrow p_{n+1}=n-p_{n+1} \Rightarrow n=2 p_{n+1} .
\end{array}
$$

Therefore, $n$ is even.
(2) Let the largest of the $n$ interior angles be $x$. Then, the angles must also include another angle $x-20^{\circ}$, and any two adjacent angles are different, with their sum not exceeding $2 x-20^{\circ}$, i.e., the average does not exceed $x-10^{\circ}$, hence
$$
\begin{array}{l}
180^{\circ}(n-2)=A_{1}+A_{2}+\cdots+A_{n} \\
\leqslant n\left(x-10^{\circ}\right), \\
n x \geqslant 180^{\circ}(n-2)+10^{\circ} n .
\end{array}
$$

Since $x > 0$, it is known that there exists a convex 34-sided polygon that meets the conditions.
From (1) to (3), it is known that the maximum value of $n$ is 34.