Something similar to Kronecker's Theorem I would like to prove (or have a reference to) the following:

Given $n$ real numbers $a_1,\ldots,a_n>0$ and $\varepsilon>0$, there exist $k,k_1,\ldots,k_n\in\mathbb N\setminus\{0\}$ such that $|ka_j-k_j|<\varepsilon$ for $j=1,\ldots,n-1$.

This is an easy consequence of Kronecker's Theorem if $a_1,\ldots,a_n,1$ are rationally independent. In this case, Kronecker says that the set $\{ka\mod\mathbb Z^n : k\in\mathbb Z\}$ is dense in $[0,1)^n$, where $a = (a_1,\ldots,a_n)\in\mathbb R^n$. In particular, there exist points close to zero in this set, which is what I am looking for.
However, if there are rational dependencies, I can't seem to find a way. Any suggestions anyone?
 A: [This answer was refering to a earlier version of the question.]
A partial solution for $n=2$. Let $\varepsilon>0$ be fixed. For simplicity denote $a:=a_1$ and $b:=a_2$. In fact, without loss of generality, one can even assume that $b=1$ by applying the injective scaling $x\mapsto x/b$.
So now we have the set $A=a\mathbb{N}+I_{\varepsilon}$ and we want to make sure that this intersects the naturals.
Case 1: If $a=p/q$ is rational, this is immediate as $qa=p\in\mathbb{N}$.
Case 2: Assume $a$ is irrational. Then by the Weil equidistribution theorem, the fractional parts $\{ka\}$ with $k\in\mathbb{N}$ are uniformly distributed on $[0,1]$. Since the measure of the interval $[0,\varepsilon]$ is strictly positive it means that one can find some infinite subset $M\subset\mathbb{N}$ such that $0\leq\{ma\}\leq\varepsilon$ for all $m\in M$, in other words it follows that $M\subset\mathbb{N}\cap A$. Clearly $M$ is unbounded.
Now for $n\geq3$, it would suffice by an inductive argument to know that $M$ contains an arithmetic progression. Indeed, if that is the case, one can just restrict the intersection of the sets for $a_1$ and $a_2$ to a set of the form $c\mathbb{N}$ and apply the same argument again.
So can an infinite set of naturals not contain an arithmetic progression of the form $c\mathbb{N}$ for some $c\in\mathbb{N}$? In general yes, for example a lacunary sequence like $1,2,4,8,16,\dots$. This occurs if and only if the gap between consecutive numbers in our set can be arbitrarily large (otherwise I think a pigeon-hole argument guarantees the existence of an arithmetic progression).
But what about our infinte set $M$? If it were to have arbitrarily large gaps between consecutive elements, that would imply that one can find arbitrarily many consecutive naturals such that $\{na\}>\varepsilon$. I suspect that this contradicts the Weil equidistribution theorem, but i am not sure.
