Slovenija 2008

a. An integer is divisible by $9$ if and only if the sum of its digits is divisible by $9$. Assume that the sum of the digits of $10^n+9n$ is divisible by $2007$. Since $2007$ is a multiple of $9$, the sum of the digits of $10^n+9n$ must be divisible by $9$. This implies that $10^n+9n$ is divisible by $9$, which is clearly not the case.

b. Let $n=1\dots 1$ (223 ones). Then $9n=9\dots 9$ (223 nines) and $10^n+9n$ is equal to $10\dots 09\dots 9$ (223 nines and $n-223=1\dots 1-223$ zeroes). The sum of the digits of this number is $1+9\cdot 223=2008$.