The sum of all the digits of the integers from 98 to 101 is

$$
9+8+9+9+1+0+0+1+0+1=38
$$

The sum of all of the digits of the integers from 1 to 2008 is
(A) 30054
(B) 27018
(C) 28036
(D) 30036
(E) 28054
First we look at the integers from 2000 to 2008.

Since we can ignore the 0s when adding up the digits, the sum of all of the digits of these integers is

$$
2+(2+1)+(2+2)+(2+3)+(2+4)+(2+5)+(2+6)+(2+7)+(2+8)=54
$$

Next, we look at the integers from 1 to 1999.

Again, since we can ignore digits of 0 , we consider these numbers as 0001 to 1999, and in fact as the integers from 0000 to 1999 , including 0000 to make 2000 integers in total.

Of these 2000 integers, 200 have a units digit of 0,200 have a units digit of 1, and so on.

(One integer out of every 10 has a units digit of 0 , and so on.)

Therefore, the sum of the units digits of these integers is

$200(0)+200(1)+\cdots+200(8)+200(9)=200+400+600+800+1000+1200+1400+1600+1800=9000$

Of these 2000 integers, 200 have a tens digit of 0,200 have a tens digit of 1 , and so on.

(Ten integers out of every 100 have a tens digit of 0 , and so on.)

Therefore, the sum of the tens digits of these integers is

$$
200(0)+200(1)+\cdots+200(8)+200(9)=9000
$$

Of these 2000 integers, 200 have a hundreds digit of 0 (that is, 0000 to 0099 and 1000 to 1099), 200 have a hundreds digit of 1 , and so on.

(One hundred integers out of every 1000 have a hundreds digit of 0 , and so on.)

Therefore, the sum of the hundreds digits of these integers is

$$
200(0)+200(1)+\cdots+200(8)+200(9)=9000
$$

Of these 2000 integers, 1000 have a thousands digit of 0 and 1000 have a thousands digits of 1 . Therefore, the sum of the thousands digits of these integers is

$$
1000(0)+1000(1)=1000
$$

Overall, the sum of all of the digits of these integers is $54+9000+9000+9000+1000=28054$.