jmc

Let $d$ be the degree of $P(x).$ Then the degree of $P(P(x))$ is $d^2.$ Hence, the degree of $P(P(x))+P(x)$ is $d^2,$ and the degree of $6x$ is 1, so we must have $d=1.$

Accordingly, let $P(x)=ax+b.$ Then $$a(ax+b)+b+ax+b=6x.$$Expanding, we get $(a^2+a)x+ab+2b=6x.$ Comparing coefficients, we get $$\begin{array}{rl}{a}^2+a& =6,\\ ab+2b& =0.\end{array}$$From the first equation, $a^2+a-6=0,$ which factors as $(a-2)(a+3)=0,$ so $a=2$ or $a=-3.$

From the second equation, $(a+2)b=0.$ Since $a$ cannot be $-2,$ $b=0.$

Hence, $P(x)=2x$ or $P(x)=-3x,$ and the sum of all possible values of $P(10)$ is $20+(-30)=\boxed{-10}.$