If $a=b$, then $(a,b,n)=(t,t,4t)$ for $t\in\Bbb N_+$, which indeed works. If $a\ne b$, WLOG let $a>b$. Since $a\mid2(a+b)$, so $a\mid2b$. But $2b<2a$, so $2b=a$, so $(a,b,n)=(2t,t,6t)$, $t\in\Bbb N_+$, which indeed works. In conclusion, $(a,b,n)=(t,t,4t)$, $(2t,t,6t)$, $(t,2t,6t)$ for $t\in\Bbb N_+$.