8. The number of positive integer solutions $(x, y, z)$ to the equation $x+y+z=2010$ that satisfy $x \leqslant y \leqslant z$ is $\qquad$ .
The equation $x+y+z=2010$ has $C_{2009}^{2}$ positive integer solutions. Among them, there is 1 positive integer solution where $x=y=z$; there are 1003 positive integer solutions where $x=y \neq z$; let there be $a$ positive integer solutions where $x<y<z$, thus $6 a+3 \cdot 1003+1=C_{2009}^{2} \Rightarrow a=335671$. There are 669 positive integer solutions where $x=y<z$, and 334 positive integer solutions where $x<y=z$, so there are $335671+669+334+1=336675$ positive integer solutions where $x \leqslant y \leqslant z$.