imc

From $|a-b|=2$ we get that $a=b\pm 2$ Similarly, $b=c\pm 3$ and $c=d\pm 4$. Substitution gives $a=d\pm 4\pm 3\pm 2$. This gives $|a-d|=|\pm 4\pm 3\pm 2|$. There are $2^3=8$ possibilities for the value of $\pm 4\pm 3\pm 2$: $4+3+2=9$, $4+3-2=5$, $4-3+2=3$, $-4+3+2=1$, $4-3-2=-1$, $-4+3-2=-3$, $-4-3+2=-5$, $-4-3-2=-9$ Therefore, the only possible values of $|a-d|$ are $9$, $5$, $3$, and $1$. Their sum is $\boxed{18}$.