So I think that the question is equivalent to tiling $\mathbb{Z}^\infty$ by translates of $A_n$. Finally, if there are $\pi(n)$ primes that are at most $n$, you might as well tile $\mathbb{Z}^{\pi(n)}$ rather than $\mathbb{Z}^\infty$. I don't know how to do this in general (unless an argument such as Victor's works), but arguably this tiling problem is a clearer view of the original question. For instance, it seems possible that the tiling can be periodic with one tile in each period, and with quotient group $\mathbb{Z}/n$. This last remark is a way to express the "linear coloring" trick in the statement of the question. As discussed in the comments, it is possible when $n+1$ or $2n+1$ is prime.

To summarize my understanding of this problem:

1. Every admissible coloring of $\mathbb{Z}_+$ yields a tiling of $\mathbb{Q}_+$ by subsets of the form $T_a = \{a,2a,3a,\ldots,na\}$, and vice versa.
2. There is a special interest in lattice tilings, in which the set of choices of $a$ is a subgroup of $\mathbb{Q}_+$ of index $n$. In particular, if $n+1$ or $2n+1$ is prime, then such a subgroup exists, and the quotient group is isomorphic to $\mathbb{Z}/n$.
3. It is easy to make variations that do not tile $\mathbb{Q}_+$, for example multiples of $\{1,2,3,4,6,8,9\}$. However, these variations generally span abelian groups in $\mathbb{Q}_+$ of lower rank than the span of $\{1,2,3,\ldots,n\}$. For the stated question, the rank is $\pi(n)$, and we may as well work in the subgroup of $P_n$ of $\mathbb{Q}_+$ generated by the $\pi(n)$ primes up to $n$.

I decided to look at the problem this way: Among all $n^{\pi(n)}$ homomorphisms from $P_n$ to $\mathbb{Z}/n$, can we heuristically estimate the number that are a bijection when restricted to $\{1,2,\ldots,n\}$? Let's say that the restriction of such a homomorphism is not particularly more likely or less likely to be a bijection than a random function. The latter probability is $n!/n^n$, so we can expect roughly $n!\cdot n^{\pi(n)-n}$ solutions. We can now take a logarithm and apply Stirling's approximation and a sufficiently careful version of the prime number theorem, $\pi(n) \approx \text{Li}(n)$. The answer is that there is plenty of entropy to have solutions; the predicted log of the number of solutions is roughly $n/(\ln n)$. This heuristic can be checked for small $n$. If the heuristic is reliable, then there should be solutions for all $n$.

It would be nice to have an effective construction of such a homomorphism for all $n$, but my impression is that the other answers so far don't find one. Note the last remark in Victor's answer: "I have a truly marvelous proof of this proposition, but the margins of MO are too thin to contain it."