Is there any order that can make complex numbers complete ordered set with least upper bound property?
I came up with $x+yi > m+ni$ if $x+y>m+n$ and if sum is equal then the complex number with greater real part is greater.
will this work? if not, are there any examples?
Your proposal does not work. For instance, the set $\{a-ai:a\in\mathbb{R}\}$ is bounded above but has no least upper bound. Its upper bounds are all numbers $x+iy$ with $x+y>0$, and there is no least such number (since, for instance, there is no least value that $x+y$ can take).
For a simple correct example, just pick any bijection $f:\mathbb{C}\to\mathbb{R}$, and define an ordering $\preceq$ on $\mathbb{C}$ by $a\preceq b$ iff $f(a)\leq f(b)$, where $\leq$ is the usual order on $\mathbb{R}$. Since the usual order on $\mathbb{R}$ is complete, so is $\preceq$.