Three, (25 points) Find the smallest positive integer $k$, such that from the 4006 numbers $1,2, \cdots, 4006$, any $k$ different numbers chosen will always include 4 numbers whose sum is 8013.
Three, from $1,2, \cdots, 4006$ take out the last 2005 numbers: $2002,2003, \cdots, 4006$,
then the sum of any four different numbers is not less than
$$
2002+2003+2004+2005=8014>8013 \text {. }
$$

Thus, $k \geqslant 2006$.
Let $x_{1}, x_{2}, \cdots, x_{2006}$ be any 2006 numbers from $1,2, \cdots, 4006$.

First, divide these 4006 numbers into 2003 pairs, each pair summing to 4007, each pair denoted as $(m, 4007-m)$, where $m=1,2$, $\cdots, 2003$.

According to the pigeonhole principle, among these 2003 pairs, there must be 3 pairs of numbers $\left(m_{1}, 4007-m_{1}\right) 、\left(m_{2}, 4007-m_{2}\right) 、\left(m_{3}, 4007-m_{3}\right)$ (where $m_{1} 、 m_{2} 、 m_{3}$ are distinct) that are 6 numbers among $x_{1}, x_{2}, \cdots, x_{2006}$.

Second, divide 4004 of these 4006 numbers (excluding 2003 and 4006) into 2002 pairs, each pair summing to 4006, each pair denoted as $(n, 4006-n)$, where $n=1,2, \cdots, 2002$.

According to the pigeonhole principle, among these 2002 pairs, there must be 1 pair of numbers $\left(n_{1}, 4006-n_{1}\right)$ that are two numbers among $x_{1}, x_{2}, \cdots, x_{2006}$.

Again, according to the pigeonhole principle, for the 1 pair of numbers $\left(n_{1}, 4006-n_{1}\right)$, at least one of the 3 pairs $\left(m_{1}, 4007-m_{1}\right) 、\left(m_{2}, 4007-m_{2}\right)$ 、 $\left(m_{3}, 4007-m_{3}\right)$, say $\left(m_{i}, 4007-m_{i}\right)$, has no common number with $\left(n_{1}, 4006-n_{1}\right)$, i.e., these 2 pairs $\left(m_{i}, 4007-m_{i}\right) 、\left(n_{1}, 4006-n_{1}\right)$ contain 4 distinct numbers, and their sum is
$$
\left[\left(m_{i}+\left(4007-m_{i}\right)\right]+\left[n_{1}+\left(4006+n_{1}\right)\right]=8013 .\right.
$$

Therefore, the smallest positive integer $k=2006$.
(Xie Wenxiao, Huanggang Middle School, Hubei, 438000)