$\tan \left(A+B\right)=\frac{\tan A+\tan B}{1-\tan A\tan B}$

$\Rightarrow$ $\frac{1}{\sqrt{3}}=\frac{y}{1-\tan A \tan B}$ where $y=\tan A+\tan B$

$\Rightarrow$ $\tan A.\tan B=1-\sqrt{3}y$

 and, also $\mathrm{AM}\ge \mathrm{GM}$

$\Rightarrow$ $\frac{\tan A+\tan B}{2} \ge \sqrt{\tan A\tan B}$

$\Rightarrow$ $y \ge 2\sqrt{1- \sqrt{3}y}$

$\Rightarrow$ $y^2 \ge 4-4\sqrt{3}y\ge 0$

$\Rightarrow$ $y \le -2\sqrt{3}-4$ or $y \ge -2\sqrt{3}+4$

$y \le -2\sqrt{3}-4$, is not possible as $\tan A+\tan B>0$

Therefore, $y \ge -2\sqrt{3}+4$

Hence, minimum positive value is $4-2\sqrt{3}$.