Solution:

Let $S(n)$ denote the sum of the digits of $n$, and let $f(x) = \sum_{n=0}^{x} S(n)$. (We may add $n=0$ because $S(0)=0$.) Observe that:

$$f(99) = \sum_{a=0}^{9} \left( \sum_{b=0}^{9} (a+b) \right) = 10 \sum_{b=0}^{9} b + 10 \sum_{a=0}^{9} a = 900$$

If $a$ is an integer between $1$ and $9$ inclusive, then:
$$\sum_{n=100a}^{100a+99} S(n) = \sum_{n=100a}^{100a+99} (a + S(n-100a)) = 100a + f(99) = 100a + 900$$

Summing, we get:
$$f(100a+99) = \sum_{n=0}^{a} (100a + 900) = 900(a+1) + 50a(a+1)$$

This formula can be used to find benchmarks. However, it turns out that this formula alone will suffice, as things turn out rather nicely:
$$\begin{aligned} 900(a+1) + 50a(a+1) &= 10000 \\ 50a^2 + 950a + 900 &= 10000 \\ 50a^2 + 950a - 9100 &= 0 \\ 50(a+26)(a-7) &= 0 \\ a &= 7 \end{aligned}$$
Therefore $f(799) = 10000$, and our answer is $799$.