4. The negation of the conclusion is easy to prove. When $a \neq b$, it must be that $(a-b) \mid [p(a)-p(b)]$, so $|a-b| \leqslant |p(a)-p(b)|$. Taking $a, b$ as $x_{i}, x_{i+1}, i=1,2, \cdots, n$ (with $x_{n+1}=x_{1}$), we get $\left|x_{1}-x_{2}\right| \leqslant \left|x_{2}-x_{3}\right| \leqslant \cdots \leqslant \left|x_{n-1}-x_{n}\right| \leqslant \left|x_{n}-x_{1}\right| \leqslant \left|x_{1}-x_{2}\right|$. Therefore, $\left|x_{1}-x_{2}\right|=\left|x_{2}-x_{3}\right|=\cdots=\left|x_{n}-x_{1}\right|$. When $n \geqslant 3$, if $i$ makes $x_{i-1}-x_{i}$ and $x_{i}-x_{i+1}$ have opposite signs, then from the above equation, we know $x_{i-1}-x_{i}=-\left(x_{i}-x_{i+1}\right)$, thus $x_{i-1}=x_{i}$, which contradicts the problem's condition. Therefore, for any $i$, we have $x_{i-1}-x_{i}=x_{i}-x_{i+1}$. At this point, if $x_{1} \geqslant x_{2}$, then $x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{n} \geqslant x_{1}$, so $x_{1}=x_{2}=\cdots=x_{n}$, which is a contradiction. If $x_{1}<x_{2}, x_{1}<x_{2}<\cdots<x_{n}<x_{1}$, it is also a contradiction. Thus, the proof is complete.