China Mathematical Competition

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 $$\begin{array}{rl}6k& =2009\times 1004-3\times 1003-1\\ & =2006\times 1005-2009+3\times 2-1\\ & =2006\times 1005-2004.\end{array}$$ We get $k=1003\times 335-334=335671$. Therefore, the number of positive integer solutions satisfying $x\le y\le z$ is $$1+1003+335671=336675.$$