6. Determine the smallest positive integer A with the following property: for any permutation of all positive integers from 1001 to 2000, it is possible to find 10 consecutive terms whose sum is greater than or equal to $A$.


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$S_{i}=b_{i}+b_{i+1}+\cdots+b_{i+9}(i=1,2, \cdots, 1991)$.
Then
$$
\begin{array}{l}
S_{1}+S_{11}+S_{21}+\cdots S_{1991} \\
=b_{1}+b_{2}+\cdots+b_{2000} \\
=\frac{(1001+2000) \times 1000}{2} \\
=1500500 .
\end{array}
$$

Thus, among $S_{1}, S_{11}, S_{21}, \cdots, S_{1991}$, there is at least one
$$
S_{i_{0}} \geqslant \frac{1500500}{100}=15005,
$$

Therefore, $A \geqslant 15005$.
Below, we construct a specific permutation to show that the smallest positive number sought is $A=15005$.

Arrange all natural numbers from 1001 to 2000 as follows:
$$
\begin{array}{l}
2000,1001,1900,1101,1800,1201, \\
1700,1301,1600,1401 ;
\end{array}
$$

Solution: Let $b_{1}, b_{2}, \cdots, b_{1000}$ be any permutation of 1001,1002,
$\cdots, 2000$.
$$
\begin{array}{l}
1999,1002,1899,1102,1799,1202, \\
1699,1302,1599,1402 ; \\
\text {... } \\
1901,1100,1801,1200,1701,1300, \\
1601,1400,1501,1300 .
\end{array}
$$

The arrangement rule is: In columns 1, 3, 5, 7, 9, each column of 100 numbers forms an arithmetic sequence with a common difference of -1; in columns 2, 4, 6, 8, 10, each column of 100 numbers forms an arithmetic sequence with a common difference of 1. The first term of each arithmetic sequence is the number in the first row. Denote the above arrangement as
$$
a_{1}, a_{2}, \cdots, a_{2000} .
$$

Let $S_{1}=a_{i}+a_{i+1}+\cdots+a_{i+9}(i=1,2, \cdots$, 1901). Then $S_{1}=15005, S_{2}=15004$. It is easy to prove by induction: If $i$ is odd, then $S_{i}=15005$; if $i$ is even, then $S=15004$. This example shows that $A$ can be 15005.
In summary, $A=15005$.
(Provided by Yan Zhenjun and Su Chun, University of Science and Technology of China)