jmc

Let $q(x)=p(x)-8.$ Then $q(x)$ has degree 4, and $q(55)=q(83)=q(204)=q(232)=0,$ so $$q(x)=c(x-55)(x-83)(x-204)(x-232)$$for some constant $c.$ Hence, $$p(x)=c(x-55)(x-83)(x-204)(x-232)+8.$$Note that $$\begin{array}{rl}p(287-x)& =c(287-x-55)(287-x-83)(287-x-204)(287-x-232)+8\\ & =c(232-x)(204-x)(83-x)(55-x)+8\\ & =c(x-55)(x-83)(x-204)(x-232)+8\\ & =p(x).\end{array}$$Hence, $p(1)=p(286),$ $p(2)=p(284),$ and so on. Therefore, $$p(1)-p(2)+p(3)-p(4)+\cdots +p(285)-p(286)=\boxed{0}.$$