A convex $n$-gon has the angles in the interval $(0,\pi)$. The equation $\tan x=0$ doesn't have solutions in this interval. The function $\tan x$ is strictly increasing on $\left(0,\dfrac{\pi}{2}\right)$. The equation $\tan x=1$ has in the interval $\left(0,\dfrac{\pi}{2}\right)$ the solution $x_1=\dfrac{\pi}{4}$. The equations $\tan x=k\in\mathbb{N},\;k\ge2$ have in the interval $\left(0,\dfrac{\pi}{2}\right)$ the solutions $x_k\in\left(\dfrac{\pi}{4},\dfrac{\pi}{2}\right)$. $\tan(\pi-\alpha)=-\tan\alpha$, hence the equations $\tan x=-k;\;k\in\mathbb{N}$ have in the interval $\left(0,\pi\right)$ the solutions $x'_k=\pi-x_k$. $\min_{k\in\mathbb{N}}\{x_k,\pi-x_k\}=x_1=\dfrac{\pi}{4};\;\max_{k\in\mathbb{N}}\{x_k,\pi-x_k\}=\pi-x_1=\dfrac{3\pi}{4}$. Let be $\beta_m=\pi-\alpha_m,\;m\in\{1,2,\dots,n-1,n\}$ the angles of the $n$-gon. Then $\sum_{m=1}^n\beta_m=(n-2)\pi\Longrightarrow \sum_{m=1}^n\alpha_m=2\pi$. $\tan\beta_m\in\mathbb{Z}\Longleftrightarrow\tan\alpha_m\in\mathbb{Z}$. $2\pi=\sum_{m=1}^n\alpha_m\ge n\cdot\dfrac{\pi}{4}\Longrightarrow n\le8$. Hence, the maximum possible value of $n$ for which the tangents of the angles of the convex $n$-gon are integer is $n_{max}=8$ and in this case, all angles have the measure $\beta_m=\dfrac{3\pi}{4},\;m\in\{1,2,\dots,8\}$.