Try to prove it!

By Fermat's Little Theorem , $10^{250} \equiv 1 \pmod{251}$ , and hence $10^{250} \equiv 1 \pmod{9\times251}$ , so that $N \; =\; \tfrac89(10^{250}-1) \; = \; 8\times251X \; = \; 2008X$ where $X$ is an integer. But $N$ is the integer whose decimal representation is a string of $250$ $8$ s.

Indeed $10^{50} \equiv 1 \pmod{251}$ , so the integer $M$ consisting of $50$ $8$ s in a row will do as well.

Consider the numbers 1, 11, 111, ... , 11...11 the last of which consists of 2009 digits 1. Since there are 2009 of these numbers, at least two of them have got to be congruent modulo 2008. The positive difference of two such congruent numbers is thus a multpile of 2008 and of the form 1...10...0 (some number of ones followed by some number of zeros). We will call this number A.

Since 2008 = 8×251, A is also divisible by 251. Furthermore, because 251 is clearly coprime to any power of 10, dividing A by an appropriate power of 10 produces a quotient of the form 11...11 which is still a multiple of 251. Multiplying this quotient by 8, we get a number with only the digit 8 in its base 10 representation that is divisible by 8×251 = 2008. The answer is yes.

We have,

$2008=251\times8$

for a number to have only 8's in its decimal representation it should be of the form

$x=8\times \dfrac{(10^{n}-1)}{9}$ where $n\geq1$

however,for it to be a multiple of $2008$ , $\dfrac{(10^{n}-1)}{9}$ must be divisible by $251$

since $251$ is prime, by Fermat's Little theorem we have,

$10^{250}\equiv 1\pmod{251}$ ,or

$10^{250}-1\equiv 0\pmod{251}$

$10^{250}-1\equiv 0\pmod{251}$

$\Rightarrow \dfrac{10^{250}-1}{9}\equiv 0\pmod{251}$ (since $251$ and $9$ are relatively prime)

thus we have,

$8\times\dfrac{10^{250}-1}{9}\equiv 0\pmod{2008}$

The answer is yes.