Three. (25 points) Given that $n$ is a positive integer. If in any permutation of $1,2, \cdots, n$, there always exists a sum of six consecutive numbers greater than 2013, then $n$ is called a "sufficient number". Find the smallest sufficient number.

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Please note that the term "温饱数" is translated as "sufficient number" to maintain the context and meaning in English. If you have a specific term you'd like to use, please let me know!
Three, the smallest well-fed number is 671.
First, prove: 671 is a well-fed number.
For any permutation $b_{1}, b_{2}, \cdots, b_{671}$ of $1,2, \cdots, 671$, divide it into 112 groups, where each group contains six numbers, and the last two groups share one number, i.e.,
$$
\begin{array}{l}
A_{1}=\left(b_{1}, b_{2}, \cdots, b_{6}\right), A_{2}=\left(b_{7}, b_{8}, \cdots, b_{12}\right), \cdots, \\
A_{111}=\left(b_{661}, b_{662}, \cdots, b_{666}\right), \\
A_{112}=\left(b_{666}, b_{667}, \cdots, b_{671}\right) .
\end{array}
$$

Then the average value of the sums of these 112 groups is greater than
$$
\frac{1+2+\cdots+671}{112}=\frac{671 \times 672}{2 \times 112}=2013 \text {. }
$$

Therefore, there must be at least one group whose sum is greater than 2013. At this point, it is always possible to find a sequence of six consecutive numbers whose sum is greater than 2013.
Thus, 671 is a well-fed number.
Next, construct a permutation of $1,2, \cdots, 670$ such that the sum of any six consecutive numbers is less than or equal to 2013.

Construct a $112 \times 6$ number table: place $670,669, \cdots, 335$ sequentially in the left $112 \times 3$ number table (larger numbers on the left (top) side of smaller numbers), and place $1,2, \cdots, 334$ sequentially in the right $112 \times 3$ number table (smaller numbers on the left (top) side of larger numbers), resulting in the number table shown in Figure 8 (with two empty positions).
\begin{tabular}{cccccc}
670 & 669 & 668 & 1 & 2 & 3 \\
667 & 666 & 665 & 4 & 5 & 6 \\
$\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ & $\vdots$ \\
340 & 339 & 338 & 331 & 332 & 333 \\
337 & 336 & 335 & 334 & &
\end{tabular}

Figure 8
Arrange the numbers in this table from left to right and from top to bottom in a single row, which will be the permutation that meets the condition (the sum of any six consecutive numbers is $2013-3k$ for $k=0,1,2,3$).
Thus, 670 is not a well-fed number.
Therefore, the smallest well-fed number is 671.
(Li Zhang, Tianjin Foreign Studies University Affiliated Foreign Languages School, 300230)