50th Mathematical Olympiad in Ukraine, Third Round (January 23, 2010)

Consider an arbitrary number. If the two last digits differ from $99$ then replacing them by $99$ we obtain a number with bigger sum of digits. Numbers $199,299,\dots ,899$ don't satisfy this condition, because for each $n\in 199,299,\dots ,899$ a number $n+100$ has the sum of digits bigger by $1$. And now, we can see that the desired number is $999$. This number has the sum of digits bigger than any other three-digit number (because all digits are $9$) and it has the sum of digits bigger than the first $100$ four-digit numbers (because the maximum sum of digits among such numbers has $1099$ and it's only $20$).