Real numbers $r$ and $s$ are roots of $p(x)=x^3+ax+b$, and $r+4$ and $s-3$ are roots of $q(x)=x^3+ax+b+240$. Find the sum of all possible values of $|b|$.
Because the coefficient of $x^2$ in both $p(x)$ and $q(x)$ is 0, the remaining root of $p(x)$ is $-(r+s)$, and the remaining root of $q(x)$ is $-(r+s+1)$. The coefficients of $x$ in $p(x)$ and $q(x)$ are both equal to $a$, and equating the two coefficients gives
\[rs-(r+s)^2 = (r+4)(s-3)-(r+s+1)^2\]from which $s = \tfrac 12 (5r+13)$. The product of the roots of $p(x)$ differs from that of $q(x)$ by $240$, so\[(r+4)\cdot \tfrac 12 (5r+7)\cdot \tfrac 12(7r+15)- r\cdot \tfrac 12 (5r+13)\cdot \tfrac 12(7r+13)=240\]from which $r^2+4r-5 =0$, with roots $r=1$ and $r=-5$.
If $r = 1$, then the roots of $p(x)$ are $r=1$, $s=9$, and $-(r+s)=-10$, and $b=-rst=90$. 
If $r = -5$, then the roots of $p(x)$ are $r=-5$, $s=-6$, and $-(r+s)=11$, and $b=-rst=-330$. 
Therefore the requested sum is $|- 330| + |90| = \boxed{420}$.