Three, (50 points) A positive integer is called "simple" if it does not have any square factors greater than 1. Determine how many numbers in $1,2, \cdots, 2010$ are simple.


Three, notice that $2010 < 45^2$.
Thus, if a number $n$ in the set $M = \{1, 2, \cdots, 2010\}$ is not simple, it must contain one or more square factors from the set of primes
$$
N = \{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43\}.
$$

Let
$$
\left[\frac{2010}{p^2}\right] = s(p), \left[\frac{2010}{p^2 q^2}\right] = s(p, q),
$$

and so on.
Then $s(2) = 502, s(3) = 223, s(5) = 80$,
$$
\begin{array}{l}
s(7) = 41, s(11) = 16, s(13) = 11, \\
s(17) = 6, s(19) = 5, s(23) = 3, \\
s(29) = s(31) = 2, \\
s(37) = s(41) = s(43) = 1.
\end{array}
$$

Using $\left[\frac{\left[\frac{a}{n}\right]}{m}\right] = \left[\frac{a}{mn}\right]$, we get
$$
\begin{array}{c}
s(2,3) = 55, s(2,5) = 20, s(2,7) = 10, \\
s(2,11) = 4, s(2,13) = 2, s(2,17) = 1, \\
s(2,19) = 1, s(2,23) = 0, s(3,5) = 8, \\
s(3,7) = 4, s(3,11) = s(3,13) = 1, \\
s(3,17) = 0, s(5,7) = 1, s(5,11) = 0; \\
s(2,3,5) = 2, s(2,3,7) = 1, \\
s(2,3,11) = s(3,5,7) = 0.
\end{array}
$$

Thus, the desired result is
$$
\begin{array}{l}
2010 - \sum_{p \in N} s(p) + \sum_{p, q \in N} s(p, q) - \sum_{p, q, r \in N} s(p, q, r) \\
= 2010 - (502 + 223 + 80 + 41 + 16 + \\
11 + 6 + 5 + 3 + 2 + 2 + 1 + 1 + 1) + \\
(55 + 20 + 10 + 4 + 2 + 1 + 1 + 8 + 4 + \\
1 + 1 + 1) - (2 + 1) \\
= 2010 - 894 + 108 - 3 = 1221.
\end{array}
$$