First, we start with the single digits. That makes 1+3+5+7=16. Then, for the two digits, we have 12 possibilities, each digit taking up 3 of each the tens and units digits. That makes (3*16*10 + 3*16)=528. For the three digit numbers, we have 4*3*2=24 possibilities. This makes 6 for each place. So we get $(6*16*111)=(96*111)=10656$. Finally, the four digit numbers. We have 24 possibilities with each digits taking 6 places. Simplifying, we get $(6*16*1111) = 106656$. Adding all of these together, we get an answer of 117,856.

fprosk wrote: Let $S$ be the set of natural numbers whose digits are different and belong to the set $\{1, 3, 5, 7\}$. Calculate the sum of the elements of $S$. Consider any place in the number. Whatever is placed here there are 3!=6 numbers with this fixed place. Now the sum for this place is (1+3+5+7)*6*10^n=16*6*10^n n being the place value. Hence the sum =16*6*(1000+100+10+1)=16*6*1111=106656