By using the addition formula $\tan(x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y}$ repeatedly, we can see that if $k$ is a positive integer, then $\tan kt$ is a rational function of $\tan t$, with integer coefficients. Thus if an equation of the type $\sum\limits_{i=1}^n \tan a_i=\tan b$ holds, then there are rational functions $A(x)$, $B(x)$ with integer coefficients such that $A(\tan 1)=B(\tan 1)$. But by the Lindemann-Weierstrass Theorem, $\tan 1$ is transcendental. It follows that $A(x)=B(x)$ identically.
The fact that $A(x)$ is identically equal to $B(x)$ implies that $\sum\limits_{i=1}^n \tan(a_it)=\tan(bt)$ for all $t$. We will show that is not possible except in the trivial case $n=1$, $a_1=b$.
For if $\sum\limits_{i=1}^n \tan(a_it)=\tan(bt)$, then $0=\lim_{t\to 0^+}\frac{\sum_{i=1}^n \tan(a_i t)-\tan(bt)}{t}=\sum_{i=1}^n a_i -b.$ But if $\sum\limits_{i=1}^n a_i=b$, and $n>1$, we cannot have $\sum_{i=1}^n \tan(a_it)=\tan(bt)$ when $t>0$ is such that the $a_it$, $bt$ are less than $\pi/2$ (in this interval, the tangent of a sum is greater than the sum of the tangents).
Added: For the more general $\sum\limits_{i=0}^n \tan a_i = \sum\limits_{j=0}^m \tan b_j$, without loss of generality no $a_i$ is a $b_j$. We argue as before that $\sum\limits_{i=0}^n \tan(a_i t) = \sum\limits_{j=0}^m \tan (b_j t)$ for all positive $t$. We can assume that $b_1 \ge b_j$ for all $j$, and that $a_i for all $i$. Let $t$ be such that $tb_1$ is very close to but below $\frac{\pi}{2}$. Then $\sum\limits_{j=1}^m \tan(b_j t)$ is very large positive, but $\sum\limits_{i=1}^n \tan(a_i t)$ is not, which gives the desired contradiction.