I claim that no such partition exists. Proof: Case 1: At least one set in the partition contains at least one odd element and at least one even element. Lemma 1. All the sets have at least one odd element and at least one even element. To see this: (Letting the sets be $ A_1,A_2,...,A_n$.) suppose $ A_1$ has an even and an odd element. Then for $ i\ge 1$, we take $ n - 1$ integers from each set besides $ A_i$, then their sum plus $ t$ must lie in $ A_i$. By changing the one integer chosen from $ A_1$, we can make this integer in $ A_i$ take on either parity, so $ A_i$ has at least one odd and at least one even integer. Lemma 2. There exists $ a$ such that $ a$ and $ a + 1$ are in different sets. This is clearly true, for otherwise one set would contain all the positive integers. Lemma 3. Let $ d = 2(a_2 + a_3 + \cdots + a_{n - 1})$, where $ a_i\in A_i$. Then if $ a_1\in A_1$, $ a_1 + d\in A_1$. To prove this, we simply have: $ a_1 + a_2 + \cdots + a_{n - 1}\in A_{n}$. $ a_2 + \cdots + a_{n - 1} + (a_1 + a_2 + \cdots + a_{n - 1}) = a_1 + d\in A_1$ Now we arrange the subsets in order $ A_1,A_2,...,A_n$ with $ a_i\in A_i$ and $ a_1'\in A_1$ satisfying: $ a_1$ is even and $ a_1'$ is odd. $ a_n = a_{n - 1} + 1$. Let $ d_1 = 2(a_2 + \cdots + a_{n - 2} + a_{n - 1})$ and $ d_2 = 2(a_2 + \cdots + a_{n - 2} + a_{n})$. Then $ d_2 = d_1 + 2$ and $ \gcd(d_1,d_2) = 1$. It is well known then that there is some $ N$ such that any even integer greater than $ N$ can be expressed as $ Xd_1 + Yd_2$ for some nonnegative integers $ X,Y$. But since $ a_1 + Xd_1 + Yd_2\in A_1$ and $ a_1' + Xd_1 + Yd_2\in A_1$ by Lemma 3, so eventually all integers are in $ A_1$. But clearly this cannot be, since if we take some high enough integer $ z\in A_1$, there will be a higher integer $ z + a_2 + a_3 + \cdots a_{n - 1}\in A_n$. Hence this is impossible, so no such partition exists. Case 2: All sets in the partition contains either only odd integers or only even integers. We can do the same thing; we just have to say that for all even (or all odd) numbers only, all even (or odd) integers starting at some point will be in $ A_1$. This is also clearly impossible. If we take $ z\in A_1$ where $ z$ is high enough, then $ z + a_2 + a_3 + \cdots a_{n - 1}\in A_n$. If $ z + a_2 + a_3 + \cdots a_{n - 1}$ is the same parity as $ z$ then this is a contradiction. Otherwise, we can show that $ A_n$ contains all odd numbers greater that a certain number. Then all integers after some point are in either $ A_1$ or $ A_n$, so we can apply the same argument and reach a contradiction.

Assume that there is such an $n$ that works. Let the sets be $A_1,\hdots, A_n$. Let $x_i$ be the least element of $A_i$. Assume with loss of generality that $1=x_1<x_2<\cdots <x_n$. The case $n=2$ is obviously impossible. If $n=3$, then $1+x_2\in A_3$. We can prove inductively that $x_2+2k\in A_2$ and $x_2+2k+1\in A_3$ for all $k\in \mathbb N$. Thus, all elements of $A_1$ are less than $x_2$. This is impossible since $A\ni 2x_2+1>x_2$. If $n\ge 4$, then we have two cases to consider $x_2=2$ and $x_2\neq 2$. In the first case, let $S=x_4+\cdots, x_n$. Therefore, $S+3\in A_3$. This implies that $2S+4\in A_2$ and $2S+5\in A_1$. We can now prove inductively that for all $k\in \mathbb N$, \begin{align*}(2k-1)S+2a+1\in A_3&\text{ if }a\in \{k,\hdots, 2k-1\}\\ 2kS+2a+1\in A_1&\text{ if }a\in \{k+1,\hdots, 2k\}\\ 2kS+2a\in A_2&\text{ if }a\in \{k+1,\hdots, 2k\}\end{align*} If $S$ is even, then \[A_3\ni (2S-1)S+2S+1=2S^2+S+1=2(S-1)S+2\cdot \frac{3S}{2}+1\in A_1\] contradiction. If $S$ is odd, then \[A_3\ni (2S-1)S+2S+1=2S^2+S+1=2(S-1)S+2\cdot \frac{3S+1}{2}\in A_2\] contradiction. Hence, this case is impossible. In the second case, again let $S=x_4+\cdots, x_n$. We have that $\{1,\hdots, x_2-1\}\subset A_1$. Therefore, $\{S+x_2+1,\hdots, S+2x_2-1\}\subset A_3$. Thus, \[A_2\ni x_2-1+S+(S+2x_2-1)=2S+3x_2-2=S+x_2+S+2x_2-2\in A_1\]contradiction. All cases lead to a contradiction so no such $n$ exists.

Suppose $A_1, A_2, \hdots, A_n$ satisfies the condition. Claim: All the subsets are eventually periodic. Proof: The condition states that for each $k$, $\sum_{i \neq k} A_i \subset A_k$. Thus, each subset contains a translate of each other subset. It follows that each subset contains a translate of itself, and is therefore eventually periodic. Since each subset contains a translate of each other subset, all subsets eventually have the same period $p$. A simple argument shows that the subsets must eventually be congruence classes modulo $p=n$, which doesn't work. Contradiction.