In a convex $n$-sided polygon, the difference between any two adjacent interior angles is $18^{\circ}$. Try to find the maximum value of $n$.

The difference between any two adjacent interior angles of a convex $n$-sided polygon is $18^{\circ}$. Try to find the maximum value of $n$.
Solution: Complete in three steps.
(1) Prove that $n$ is even. Let the convex $n$-gon be $A_{1} A_{2} A_{3} \cdots$
$A_{n}$, then $A_{2}=A_{1} \pm 18^{\circ}$, (where “ + ” or “-” is chosen)
$A_{3}=A_{2} \pm 18^{\circ}=A_{1} \pm 18^{\circ} \pm 18^{\circ}$, (choose “ + ” or “ - ” as before) $\qquad$
$$
\begin{array}{l}
A_{k}=A_{1} \pm \underbrace{18^{\circ} \pm 18^{\circ} \pm \cdots \pm 18^{\circ}}_{\text {total }} \\
=A_{1}+p_{k} \cdot 18^{\circ}-q_{k} \cdot 18^{\circ} .
\end{array}
$$

where $p_{k}$ is the number of “ + ” signs, and $q_{k}$ is the number of “ - ” signs, so $p_{k}+q_{k}=k-1$.
In the above formula, let $k=n+1$, then we know
$$
\begin{aligned}
A_{n+1} & =A_{1}+p_{n+1} \cdot 18^{\circ}-q_{n+1} \cdot 18^{\circ} \\
& =A_{1}+p_{n+1} \cdot 18^{\circ}-\left(n+1-1-p_{n+1}\right) \cdot 18^{\circ} \\
& =A_{1} .
\end{aligned}
$$
(Actually, vertex $A_{n+1}$ is the same as $A_{1}$)
Thus, $p_{n+1}=n-p_{n+1}$,
which means $n=2 p_{n+1}$, proving that $n$ is even.
(2) Prove that $nn x \geqslant(n-2) \cdot 180^{\circ}+n \cdot 9^{\circ} \text {. }$
$$

Then $n0$,
indicating that there exists a convex 38-gon that meets the conditions.
In summary, the maximum value of $n$ is 38.
(Yang Yongliang and Xun Yangtao, No. 5 Middle School of Bozhou City, Hunan Province, 414411)