jmc

The cubic passes through the points $(2,1),$ $(7,19),$ $(15,11),$ and $(20,29).$ When these points are plotted, we find that they form the vertices of a parallelogram, whose center is $(11,15).$ We take advantage of this as follows.



Let $f(x)=p(x+11)-15.$ Then $$\begin{array}{rl}f(-9)& =p(2)-15=-14,\\ f(-4)& =p(7)-15=4,\\ f(4)& =p(15)-15=-4,\\ f(9)& =p(20)-15=14.\end{array}$$Now, let $g(x)=-f(-x).$ Then $$\begin{array}{rl}g(-9)& =-f(9)=-14,\\ g(-4)& =-f(4)=4,\\ g(4)& =-f(-4)=-4,\\ g(9)& =-f(-9)=14.\end{array}$$Both $f(x)$ and $g(x)$ are cubic polynomials, and they agree at four different values, so by the Identity Theorem, they are the same polynomial. In other words, $$-f(-x)=f(x).$$Then $$15-p(11-x)=p(x+11)-15,$$so $$p(11-x)+p(x+11)=30$$for all $x.$

Let $$S=p(1)+p(2)+p(3)+\cdots +p(21).$$Then $$S=p(21)+p(20)+p(19)+\cdots +p(1),$$so $$2S=[p(1)+p(21)]+[p(2)+p(20)]+[p(3)+p(19)]+\cdots +[p(21)+p(1)].$$Since $p(11-x)+p(x+11)=30,$ each of these summands is equal to 30. Therefore, $$2S=21\cdot 30=630,$$and $S=630/2=\boxed{315}.$