smc

We start off with the fact that the median is $9$, so we must have $a,b,c,d,e,9,f,g,h,i,j$, listed in ascending order. Note that the integers do not have to be distinct. Since the mode is $8$, we have to have at least $2$ occurrences of $8$ in the list. If there are $2$ occurrences of $8$ in the list, we will have $a,b,c,8,8,9,f,g,h,i,j$. In this case, since $8$ is the unique mode, the rest of the integers have to be distinct. So we minimize $a,b,c,f,g,h,i$ in order to maximize $j$. If we let the list be $1,2,3,8,8,9,10,11,12,13,j$, then $j=11\times 10-(1+2+3+8+8+9+10+11+12+13)=33$. Next, consider the case where there are $3$ occurrences of $8$ in the list. Now, we can have two occurrences of another integer in the list. We try $1,1,8,8,8,9,9,10,10,11,j$. Following the same process as above, we get $j=11\times 10-(1+1+8+8+8+9+9+10+10+11)=35$. As this is the highest choice in the list, we know this is our answer. Therefore, the answer is $\boxed{35}$