Three, (25 points) In any permutation of $1,2, \cdots, 200$, there always exists a sum of 20 consecutive numbers that is not less than $a$. Find the maximum value of $a$.


Three, take any permutation, and divide it into 10 groups, each containing 20 numbers. Let $A_{1}=\left(b_{1}, b_{2}, \cdots, b_{20}\right), A_{2}=\left(b_{21}, b_{2}\right.$, $\left.\cdots, b_{40}\right), \cdots, A_{10}=\left(b_{181}, b_{182}, \cdots, b_{200}\right)$.
Take the average of the 10 groups of numbers
$$
\bar{A}=\frac{1+2+\cdots+200}{10}=2010 \text {. }
$$

Obviously, at least one of $A_{1}, A_{2}, \cdots, A_{10}$ is not less than $\bar{A}$. At this point, it is always possible to find a sum of 20 consecutive numbers that is not less than 2010.

Below, construct a permutation of $1, 2, \cdots, 200$ such that the sum of any 20 consecutive numbers is not greater than 2010.

Construct a $10 \times 20$ number table: place 200, 199, $\cdots, 101$ sequentially in the first $10 \times 10$ number table (large numbers on the left (top) side of smaller numbers), and place $1, 2, \cdots, 100$ sequentially in the second $10 \times 10$ number table (small numbers on the right (top) side of larger numbers).

Arrange the numbers in this table from left to right and from top to bottom into a sequence, which is the permutation that meets the requirements of the problem (the sum of any 20 consecutive numbers is one of $2010-10 i(i=0,1, \cdots, 10)$).
Thus, the maximum value of $a$ is 2010.
Note: The constructive part of the proof in this problem was provided by Mr. Song Qiang from our editorial department.