For $n \in \mathbf{N}_{+}$, define
$$
S(n)=\left[\frac{n}{10^{[18 n]}}\right]+10\left(n-10^{[i \mid n]}\left[\frac{n}{10^{\left[1 / B^{n}\right]}}\right]\right),
$$

where $[x]$ denotes the greatest integer not exceeding the real number $x$. Then, among $1,2, \cdots, 2012$, the number of positive integers $n$ that satisfy $S(S(n))=n$ is
6. 108.

Let $t=10^{[\lg n]}$. Then
$$
S(n)=\left[\frac{n}{t}\right]+10\left(n-t\left[\frac{n}{t}\right]\right) \text {. }
$$

Notice that, $n-t\left[\frac{n}{t}\right]$ is the remainder of $n$ modulo $t$, and $\left[\frac{n}{t}\right]$ is the first digit of $n$.
We will discuss the cases below.
(1) If $n$ is a one-digit number, all satisfy the requirement, totaling 9.
(2) If $n=\overline{x y}, S(n)=\overline{y x}, S(S(n))=\overline{x y}$, totaling 81 (excluding those with $y=0$).
(3) If $n=\overline{x y z}, S(n)=\overline{y z x}, S(S(n))=\overline{z x y}$, thus, $x=y=z$, totaling 9.
(4) If $n=\overline{x y z w}, S(S(n))=\overline{z w x y}$, thus, $w=y, z=x$, totaling 9.
Therefore, the number of $n$ that satisfy the requirement is 108.