Most of this problem is simple chains of deduction. By noting that SIX + TEN = SIXTEEN, we see that TEN = TEEN and therefore E must equal 0, and we can remove all E's from consideration. We can also subtract 10 and TN from each of the -teens, to get such results as IGHT = IGH (and therefore T=0), FIV = FIF (and therefore F=V), and so on.
Even though H,R,U,G are left undetermined, the sum of all the letters can still be found! In solving, we obtain g+h=19; h+r=14; r+u=5. Then, by adding the first and last equations, we get g+h+r+u=26, and all the missing values are accounted for. Add this to the known values and the overall sum is 50.