Question 233: Let the difference between any two adjacent interior angles of a convex $\mathrm{n}$-sided polygon be $20^{\circ}$, find the maximum possible value of $\mathrm{n}$.
Question 233, Solution: Let the convex $n$-sided polygon satisfying the conditions be $A_{1} A_{2} \ldots A_{n}$.
Let the exterior angle of $A_{i}(1 \leq i \leq n)$ be $\alpha_{i}$, then $\sum_{i=1}^{n} \alpha_{i}=360$, and $\alpha_{i+1}-\alpha_{i} \in\{20,-20\}$, where $1 \leq i \leq n$, and $\alpha_{n+1}=\alpha_{1}$.
Let $\alpha_{i+1}-\alpha_{i}=k_{i} \cdot 20^{\circ}$, where $k_{i} \in\{-1,1\}$, then
$$
\begin{array}{l}
0=\sum_{i=1}^{n}\left(\alpha_{i+1}-\alpha_{i}\right)=\left(\sum_{i=1}^{n} k_{i}\right) \cdot 20^{\circ} \\
\Rightarrow \sum_{i=1}^{n} k_{i}=0
\end{array}
$$

Thus, the number of 1s and -1s in $k_{1}, k_{2}, \ldots, k_{n}$ is the same, so $n$ is even. Let $n=2 t$.
According to the conditions, $\alpha_{2i-1}+\alpha_{2i}>20^{\circ}(1 \leq i \leq t)$. Therefore,
$$
360^{\circ}=\sum_{i=1}^{t}\left(\alpha_{2 i-1}+\alpha_{2 i}\right)>20 t^{\circ}
$$

Thus, $t<18$, which means $t \leq 17$. Therefore, $n=2 t \leq 34$.
On the other hand, when $n=34$, take $\alpha_{1}=\alpha_{3}=\ldots=\alpha_{33}=\frac{10}{17}^{\circ}, \alpha_{2}=\alpha_{4}=\ldots=\alpha_{34}=\frac{350}{17}^{\circ}$, which satisfies the conditions.
In summary, the maximum value of $n$ that satisfies the conditions is 34.