Three. (50 points) In a certain drill activity, the drill leader arranged $n$ students, numbered $1 \sim n (n>3)$, in a circular formation, and they performed a $1-2-3$ cyclic counting. The drill leader recorded the numbers of the students who reported, and required students who reported 1 and 2 to leave the formation, while students who reported 3 remained in the formation, and their numbers were changed to the sum of the numbers of the three students in this cycle of counting. The counting continued in this manner. When the number of students remaining on the playground was no more than two, the counting activity ended. The drill leader recorded the numbers of the students who remained on the playground (for example, if nine students numbered $1 \sim 9$ were arranged in a circular formation, after the counting ended, only the student originally numbered 9 remained on the playground, at which point his number was 45, and the data recorded by the drill leader were $1,2,3,4,5,6,7,8,9,6,15,24,45$). It is known that 2011 students participated in the drill.
(1) What was the initial number of the student who remained on the playground?
(2) Find the sum of the numbers recorded by the drill leader.
Three, the numbers recorded by the leader are $a_{1}$, $a_{2}, \cdots, a_{n}, a_{n+1}, \cdots, a_{m}$, with their sum being $S_{n}$. Thus,
(1) After each 1-2-3 counting cycle, the total number of students decreases by 2, but the sum of the numbers remains unchanged.
(2) If $n=3^{k}$, after $3^{k-1}$ 1-2-3 counting cycles, $3^{k-1}$ new numbers are generated, and the sum of these numbers is $\frac{(1+n) n}{2}$; after another $3^{k-2}$ counting cycles, $3^{k-2}$ new numbers are generated, and the sum of these numbers is also $\frac{(1+n) n}{2}; \cdots \cdots$ After a total of
$$
3^{k-1}+3^{k-2}+\cdots+3^{1}+1=\frac{3^{k}-1}{2}
$$
counting cycles, only one student remains, and his number is $\frac{(1+n) n}{2}$, and his original number is $n$ (i.e., when $n=3^{k}$, the last student remains on the field).
From (1), we know $S_{n}=(k+1) \frac{(1+n) n}{2}$.
(3) If the number of students is $3 k$, and the number of the last student is $a_{p}$, then after $k$ counting cycles, $k$ new numbers are generated, and the number of students becomes $k$, with the first student's number being $a_{p+1}$.
For a general odd number $n(n>3)$, let $k$ and $r$ satisfy $n=3^{k}+2 r\left(0 \leqslant r<3^{k}\right)$.
Then $3^{k}<n<3^{k+1}, 3 r<n$. One group, the sum of their numbers is
$$
M=1+2+\cdots+3 r=\frac{3 r(3 r+1)}{2} .
$$

Let the initial sum of student numbers be $N=\frac{n(n+1)}{2}$.
From (1) and (3), we know that after $r$ 1-2-3 counting cycles, the number of students becomes $3^{k}$, with the first student's number being $a_{3 r+1}$, called the second group; after another $3^{k-1}$ counting cycles, the number of students becomes $3^{k-1}$, called the third group; repeat the counting cycles until the number of students is 1, which is the $k$ +2 group.
The original number of the last student remaining on the field is $3 r$, and
$$
\begin{array}{l}
S_{n}=M+(k+1) N \\
=\frac{3 r(3 r+1)}{2}+(k+1) \frac{(1+n) n}{2} .
\end{array}
$$

When $n=2011$, since $n=2011=3^{6}+2 \times 641$, we have $k=6, r=641$.
The original number of the last student remaining on the field is 1923. Therefore, $S_{2011}=16011388$.