Can you just simply add numbers?

There are $5$ ways to choose the first digit, $5$ ways to choose the $2$ nd digit, $5$ ways to choose the third digit and $5$ ways to choose the fourth digit.

There are $5^4=625$ $4$ -digit numbers having only odd digits.

Each of $1,3,5,7,9$ occurs $125$ times.

Therefore the sum of all the units digits is $125\times (1+3+5+7+9) = 125\times (25) = 3125$ .

Sum of tens digits is $3125\times 10$

Sum of hundreds and thousands digits are respectively $3125\times 100$ and $3125 \times 1000$

Sum of all such 4 digit numbers = $3125+3125\times 10+3125\times 100+3125\times 1000=\\ \boxed{3471875}\\$

Generalization for n digit numbers

In this case we need to find the sum of all n-digit numbers ,all the digits of which are odd.

There are $5^n$ $n$ -digit numbers having the desired property.

Each of $1,3,5,7,9$ occurs $\frac{5^n}{5}=5^{n-1}$ times

Sum of units digit = $5^{n-1}\times(1+3+5+7+9)=5^{n+1}$

Similarly we can do for tens, hundreds upto $10^{n-1}$ .

Answer= $5^{n+1}+10\times5^{n+1}+100\times5{n+1}+...+10^{n-1}\times5^{n+1}$ and we are done

Since all the digits have to be odd, we are limited to 1, 3, 5, 7 and 9 as digits.

There are 5 possibilities for each digit which means there are totally 625 possible numbers. The average of each digit of each number is 5 which means that the average of all these odd digit numbers is 625.

Average=Sum/Number of items
5555=x/625
x=625*5555
=3471875

We have $5^{3}$ ways to vary the another digits except for the one we consider.

Since all digit is an odd number, so, the sum is $1+3+5+7+9=5^{2}$ and is the same in all digits.

Hence, we have $5^{5} \times 1111=\boxed{3471875}$