I spent an excessive amount of time thinking about this problem under the assumption that the answer was "no", but in fact there's a very simple partition with the desired property. The number 12 is significant only in that $12 + 1$ is prime. The twelve desired subsets are defined as follows: $S_i$ consists of those integers which, when written in base 13, have rightmost nonzero digit equal to $i$. Thus, 1, 13, 14, 27, and 169 all belong to $S_1$, while 2, 15, 26, and 28 all belong to $S_2$. This partition has the desired property because the numbers 1, 2, ..., 12 form a group under multiplication modulo 13. This leaves open the interesting question of what happens if we replace 12 with an integer not one less than a prime. For example, is it possible to partition the integers into three parts so that for every $k$, the numbers $k$, $2k$ and $3k$ lie in different parts?

Edit/Update: okay, so three was too small. Solution

But what about with five? Seven? Twenty-seven?

Thanks for sharing your neat solution, JBL. I thought the answer was "no" when I posted this problem. Consequently, it appears I might have posted in the wrong forum. Should this be moved to "Number Theory"?

Well, it's both combinatorics and number theory; as it happens, I'm a moderator, so I can move it Incidentally, the solution I gave for 3 also works for 12, and can generate the same partition (as well as others): given $n = 2^a \cdot 3^b \cdot 5^c \cdot 7^d \cdot 11^e \cdots$, put $n$ in the set $S_i$ where $i \equiv a + 4b + 9c + 11d + 7e \pmod{12}$. In fact, solutions like this work for all "small" values of the parameter 12 (say, up to at least 14 or 15 and probably higher than this). Can they really work for all integers?

I don't think it's a question of smallness; you can, for instance, get a solution of that form for any number of the form $p-1$, for prime $p$. If we want to map $p_1^{e_1}p_2^{e_2}\cdots$ to a number $a_1 e_1 + a_2 e_2 + \cdots \pmod {p-1}$, we can pick the coefficients $a_i$ as follows: let g be any primitive root mod p, then choose $a_i$ such that $g^{a_i} \equiv p_i \mod p$. For instance, this gives your example of $a + 4b + 9c + 11d + 7e\pmod {12}$ for the primitive root $g=2\pmod {13}$. Unfortunately, I still don't see any way that this could be extended to numbers one less than non-primes.

$12$ can be change to any integer $a>1$: The positive integers be partitioned into $a$'s subset ${S_1},{S_2},...,{S_a}$, such that for each positive integer $k$, the numbers $k,2k,...,ak$ belong to different subsets. And we can prove it without any advanced knowledge like "primitive root". Proof: We put $1$ to $S_1$, if $1,2,...,n$ are proper arranged, we can put $n+1$ in some $S_i$, which without any of $\frac{{n + 1}}{2},\frac{{n + 1}}{3},...,\frac{{n + 1}}{a}$ in it. So we can arrange any positive integer, and we done.

yunxiu wrote: Proof: We put $1$ to $S_1$, if $1,2,...,n$ are proper arranged, we can put $n+1$ in some $S_i$, which without any of $\frac{{n + 1}}{2},\frac{{n + 1}}{3},...,\frac{{n + 1}}{a}$ in it. So we can arrange any positive integer, and we done. Wrong proof. For instance let $a=7$. Assume that $1, 2, \ldots, 2099$ are placed, where do you put 2100? It can't go to the subsets of $300,600,...,1800$, neither can it go into the subsets of $525,1050, 1575$, etc... There are too many restrictions, you can't be sure, that a suitable subset can be found.

I think this proof is valid for every $n=2^x3^y5^z7^u11^vw,(w,2310)=1$,if $7x+4y+3z+2u+v\equiv k(mod12)$,then put n in the k-th set.

furthermore,I think that this problem has some relation with Rado's theorem?

I have a nice solution, if it is true. Write all positive integers in form $13^{a}*b$ such that $13$ doesn't divide $b$, then divide positive integers to $12$ groups such that, if $n=13^{m}*t$ (such that $13$ doesn't divide $t$) and if $t \equiv x \pmod{13}$, then put $n$ to $x^{th}$ group ($x \in (1,2,3,4,5,6,7,8,9,10,11,12)$). It is easy to see that such dividing satisfies the given condition.