Let $n = \overline{d_1d_2d_3d_4} = 1000d_1 + 100d_2 + 10d_3 + d_4$, $d_1, d_2, d_3, d_4 \in [0, 1, 2, \dots, 9]$, and $S(n) = d_1 + d_2 + d_3 + d_4$.
Match the non-negative multiples of 6 less than 2000 into 167 pairs $(x, y)$, $x + y = 1998$, such that
$(0, 1998), (6, 1992), (12, 1986), \dots, (996, 1002).$
For each pair $(x, y)$, let $x = \overline{a_1a_2a_3a_4}$, $y = \overline{b_1b_2b_3b_4}$, then
$$1000(a_1 + b_1) + 100(a_2 + b_2) + 10(a_3 + b_3) + (a_4 + b_4) = x + y = 1998.$$
Since $x, y$ are even, $a_4, b_4 \le 8$. So $a_4 + b_4 \le 16 < 18$. Thus, $a_4 + b_4 = 8$. Since $a_3 + b_3 \le 18 < 19$, $a_3 + b_3 = 9$.
Similarly, we can obtain $a_2 + b_2 = 9$ and $a_1 + b_1 = 1$. Thus,
$$\begin{aligned} S(x) + S(y) &= (a_1 + b_1) + (a_2 + b_2) + (a_3 + b_3) + (a_4 + b_4) \\ &= 1 + 9 + 9 + 8 = 27. \end{aligned}$$
Consequently, there is only one of $S(x)$ and $S(y)$ that is the multiple of 6. (This is because that $x$, $y$ are all multiples of 3, so are $S(x)$ and $S(y)$.) That is, there is only one of $x$ and $y$ which is a six match number.
Therefore, there are 167 six match numbers less than 2000, and there is just one six match number between 2000 and 2011.
Therefore, the answer is $167 + 1 = 168$.