The numbers $n$ and $S(n)$ have the same remainder when divided by $3$ and $9$. Namely, if $n = \overline{a_m a_{m-1} \dots a_1 a_0}$ then
$$\begin{aligned} n &= \overline{a_m a_{m-1} \dots a_1 a_0} = 10^m a_m + 10^{m-1} a_{m-1} + \dots + 10 a_1 + a_0 \\ &= [(10^m - 1)a_m + (10^{m-1} - 1)a_{m-1} + \dots + (10 - 1)a_1] + [a_m + a_{m-1} + \dots + a_1 + a_0] \end{aligned}$$
Each expression in the brackets is divisible by $9$, therefore $n$ and $S(n)$ have the same remainder when divided by $3$ and $9$. The numbers $n$, $S(n)$, $S(S(n))$ have the same remainder when divided by $3$. Therefore the sum $n + S(n) + S(S(n))$ is always divisible by $3$ while $2011$ is not. Therefore the equation has no solution.