jmc

Let $f(x)=p(x)-17x.$ Then $f(1)=f(2)=f(3)=0.$ Also, $f(x)$ is a monic polynomial of degree 4, so $$f(x)=(x-1)(x-2)(x-3)(x-r),$$for some real number $r.$ Then $$p(x)=f(x)+17x=(x-1)(x-2)(x-3)(x-r)+17x.$$Therefore, $$\begin{array}{rl}p(0)+p(4)& =(0-1)(0-2)(0-3)(0-r)+17\cdot 0+(4-1)(4-2)(4-3)(4-r)+17\cdot 4\\ & =6r+24-6r+68\\ & =\boxed{92}.\end{array}$$