120. In a convex $n$-sided polygon, the difference between any two adjacent interior angles is $1^{\circ}$. Find the maximum value of the difference between the largest and smallest interior angles.
Solution: Let the largest interior angle be $\alpha$, and the smallest interior angle be $\alpha-k$ (omitting the degree symbol for simplicity, the same applies below). Therefore, what we are looking for is the maximum value of $k$. Hence, the interior angles should also include $\alpha-1, \alpha-2, \alpha-3, \cdots$, $\alpha-(k-1)$, etc. Since the process of changing from $\alpha$ to $\alpha-k$ can be done in two directions, clockwise and counterclockwise, the above $k-1$ data points should each appear at least twice.

In addition, the other $n-2k$ interior angles $\beta_{1}, \beta_{2}, \cdots, \beta_{n-2k}$ should also take values from the above $k+1$ data points. Suppose that among the aforementioned $2k$ angles, there are $p$ angles $\left.\mid \beta_{1}\right\}(i=1,2, 3, \cdots, n-2k)$ inserted between two adjacent interior angles $\alpha_{1}$ and $\alpha_{2}$. Assume these are $\beta_{1}, \beta_{2}, \beta_{3}, \cdots, \beta_{p}$. Since $[x \pm 1]=[x] \pm 1$ (where [] is the notation for the floor function), the parity of the integer parts of any two adjacent interior angles is exactly different.

In the above changes, $\left[\alpha_{1}\right],\left[\beta_{1}\right],\left[\beta_{2}\right],\left[\beta_{3}\right], \cdots,\left[\beta_{p}\right]$, $\left[\alpha_{2}\right]$ alternate in parity, and since $\left[\alpha_{2}\right]=\left[\alpha_{1} \pm 1\right]$ has a different parity from $\left[\alpha_{1}\right]$, $p$ should be even. At the same time, since no two adjacent angles are the same, the sum of the two angles should not exceed $\alpha+(\alpha-1)$, so the sum of these $p$ angles should not exceed $\frac{p}{2}[\alpha+(\alpha-1)]$. Clearly, the same property applies to the other angles, meaning the sum of these $n-2k$ angles should not exceed $\frac{n-2k}{2}[\alpha+(\alpha-1)]$.
By the interior angle sum theorem, we have
$$
\begin{array}{l}
(n-2) \cdot 180 \\
= \alpha+2[(\alpha-1)+(\alpha-2)+(\alpha-3)+\cdots \\
\quad+(\alpha-(k-1))]+(\alpha-k)+\left(\beta_{1}+\beta_{2}+\beta_{3}+\right. \\
\left.\cdots+\beta_{n-2 k}\right) \\
= \alpha+2 \times \frac{(k-1)[\alpha-1+\alpha-(k-1)]}{2} \\
\quad+\alpha-k+\left(\beta_{1}+\beta_{2}+\cdots+\beta_{n-2 k}\right) \\
\leqslant 2 k \alpha-k^{2}+\frac{n-2 k}{2}[\alpha+(\alpha-1)] .
\end{array}
$$

Rearranging and combining with $\alpha > 0$,
we know that a convex polygon satisfying $k_{\max }=18$ exists.
In summary, the maximum value sought is $18^{\circ}$.