It is easy to find that the number of positive integer solutions of $x + y + z = 2010$ is $C_{2009}^2 = 2009 \times 1004$.
We now classify these solutions into three categories:

(1)
$x = y = z$, the number in this category is obviously 1;

(2) there are exactly two that are equal among $x, y, z$ —
the number in this category is 1003;

(3) $x, y, z$ are different from each other — suppose the number in this category is $k$.
From
$$1 + 3 \times 1003 + 6k = 2009 \times 1004,$$
we have
$$\begin{align*} 6k &= 2009 \times 1004 - 3 \times 1003 - 1 \\ &= 2006 \times 1005 - 2009 + 3 \times 2 - 1 \\ &= 2006 \times 1005 - 2004. \end{align*}$$
We get $k = 1003 \times 335 - 334 = 335671$.
Therefore, the number of positive integer solutions
satisfying $x \le y \le z$ is
$$1 + 1003 + 335671 = 336675.$$