This is not a solution, but ideally a clarification of the question. The semigroup $\mathbb{Z}_+$ under addition is equivalent to the semigroup $\mathbb{Z}_{\ge 0}^\infty$ under addition. We are interested in certain special subsets $A_n$ of the latter, which in multiplicative form correspond to $\{1,2,\ldots,n\}$. Each $A_n$ is a certain $n$-element lower set in the natural partial ordering on $\mathbb{Z}_{\ge 0}^\infty$. We would like to color $\mathbb{Z}_{\ge 0}^\infty$ with $n$ colors such that each translate of $A_n$ uses $n$ different colors.
By a compactness argument, we can find a sequence of translates of $\mathbb{Z}_{\ge 0}^\infty$ in $\mathbb{Z}^n$ such that the colorings converge to a coloring of all of $\mathbb{Z}^\infty$. In other words, in multiplicative form, we could equally well have asked the question in $\mathbb{Q}_+$ as in $\mathbb{Z}_+$.
Also, I think that the subset of $\mathbb{Z}^\infty$ with color 1 is a set of vectors that yields a tiling of $\mathbb{Z}^\infty$ by translates of $-A_n$. For instance, if two of these translates intersect at $x$, then you correspondingly $x+A_n$ contains two elements with color 1. (Technically this only shows that it is a packing, not a tiling. However, the cracks that are not tiled must have density 0, and you can translate them away by the same compactness argument.) Moreover, if you have such a tiling, then you can color each element of each tile in the same way, and then for that coloring each translate $x+A_n$ has all different colors. Again, in multiplicative form, you're tiling the positive rational numbers by arithmetic progressions $a,2a,3a,\ldots,na$.
To summarize my understanding of this problem:
- 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.
- 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 $\mathbb{Z}/n$.
- 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."