Sum of irrational numbers, a basic algebra problem

This is and old question and the answer is yes.

Mordell was interested in a particular case of this problem in his paper On the linear independence of algebraic numbers.

Later on, Ursell in the paper The degrees of radical extensions solves this problem in the particular case $x_i\in\mathbb N^*$ and $(x_i,x_j)=1$ for $i\neq j$.

In fact, the problem can be easily reduced to the case where $x_i\in\mathbb N^*$ and $l_1=\cdots=l_n$. This is in fact a problem proposed/solved by Preda Mihailescu long time go. Later on Toma Albu worked on this problem from different point of view and proved this as Theorem 2.2 in this survey paper. Basically he proved the theorem in an older paper where he acknowledges that his results have appeared in many other sources, such as Karpilovski's book Field Theory.

Anyway, the proof given by Mihailescu is the most "elementary" and I suggest you to try to read it (well, maybe using Google translate).

Too long for a comment.

Your problem is actually equivalent to the following:

Problem: If $a_1,..,a_k$ are positive integers, and $m$ is a positive integer, so that $\sqrt[m]{a_i} \notin \mathbb Q$ then

$$\sum \sqrt[m]{a_i} \notin \mathbb Q \,$$

You can do the reduction to this problem in two steps:

Step 1: Let $m=lcm (l_1,..,l_n)$. Then $\frac{1}{l_i}=\frac{k_i}{m}$. Let $y_i=x_i^{k_i}$.

Then, you know that $y_i$ are positive rational numbers and $x_i^\frac{1}{l_i}=\sqrt[m]{y_i}$.

Step 2: Let $y_i=\frac{b_i}{c_i}$ and let $d= lcm (c_1,..,c_n)$. Then you can write

$$y_i=\frac{a_i}{d^m}$$

and you get that problem.

I am pretty sure that I saw some Theorems in Galois Theory which imply stronger versions of the above problem.

Added Theorem 2.9 in This Paper is exactly the question asked, it provides some reference to the History of the problem and a proof.