Let us solve in positive integers $1\leq x\leq y\leq z$, with $x,y,z \mid x+y+z$. It follows $z \mid x+y \leq 2z$, so (1) either $x+y = 2z$, thus $x=y=z$, with solutions $x(1,1,1)$; (2) or $x+y = z$, but then $y \mid z+ x = y+2x$, thus $y \mid 2x \leq 2y$, so (2.1) either $x=y$, with solutions $x(1,1,2)$; (2.2) or $y=2x$, with solutions $x(1,2,3)$. Back to our ordered triplets, we have $2010$ for point (1), $3\cdot 1005 = 3015$ for point (2.1), and $6\cdot 670 = 4020$ for point (2.2); in total $9045$ ordered triplets.