Let $n$ be a positive integer. We call $(a_1,a_2,\cdots,a_n)$ a [i]good[/i] $n-$tuple if $\sum_{i=1}^{n}{a_i}=2n$ and there doesn't exist a set of $a_i$s such that the sum of them is equal to $n$. Find all [i]good[/i] $n-$tuple.
(For instance, $(1,1,4)$ is a [i]good[/i] $3-$tuple, but $(1,2,1,2,4)$ is not a [i]good[/i] $5-$tuple.)
1. **Assumption and Initial Setup:**
   We assume that \(a_i\) is a positive integer for all \(1 \leq i \leq n\). Let \(a_1 \leq a_2 \leq \cdots \leq a_n\). We need to find all *good* \(n\)-tuples \((a_1, a_2, \ldots, a_n)\) such that \(\sum_{i=1}^{n} a_i = 2n\) and no subset of \(a_i\)'s sums to \(n\).

2. **Sum Constraints and Modulo Analysis:**
   Consider the partial sums \(a_1, a_1 + a_2, \ldots, a_1 + a_2 + \cdots + a_{n-1}\). These sums are all between \(0\) and \(2n\) exclusive. If any of these sums is congruent to \(0 \pmod{n}\), it would equal \(n\), which contradicts the condition that no subset sums to \(n\). Additionally, if any two partial sums are congruent modulo \(n\), their difference would also be a sum of some \(a_i\)'s and would equal \(n\), another contradiction. Therefore, all these sums must be distinct and nonzero modulo \(n\).

3. **Induction Hypothesis:**
   We use strong induction to show that \(a_1 \equiv a_2 \equiv \cdots \equiv a_n \pmod{n}\). For the base case, consider \(a_2 \equiv a_1 + \cdots + a_k \pmod{n}\) for some \(1 \leq k \leq n\). If \(k \geq 2\), we get a contradiction. Thus, \(k = 1\).

4. **Inductive Step:**
   Assume \(a_1 \equiv a_2 \equiv \cdots \equiv a_k \pmod{n}\) for \(1 \leq k < n\). Consider the sum \(a_2 + \cdots + a_{k+1} \equiv a_1 + \cdots + a_{k-1} + a_{k+1} \pmod{n}\). If \(i \leq k-1\), we get a contradiction. If \(i \geq k+2\), we get another contradiction. If \(i = k+1\), then from the previous step, we have \(a_k \equiv 0 \pmod{n}\). So, \(i = k\), and we conclude \(a_{k+1} \equiv a_1 \pmod{n}\). Therefore, \(a_1 \equiv \cdots \equiv a_n \pmod{n}\).

5. **Form of \(a_i\):**
   We can write \(a_1 = k_1 \times n + r, \ldots, a_n = k_n \times n + r\) where \(0 < r < n\). Thus, \(r + k_1 + \cdots + k_n = 2\).

6. **Case Analysis:**
   - **Case 1:** If \(r = 2\), then \(k_1 + \cdots + k_n = 0\). This implies \(a_1 = a_2 = \cdots = a_n = 2\), which is a solution.
   - **Case 2:** If \(r = 1\), then \(k_1 + \cdots + k_n = 1\). Given \(k_1 \leq k_2 \leq \cdots \leq k_n\), we have \(k_1 = k_2 = \cdots = k_{n-1} = 0\) and \(k_n = 1\). This gives the solution \(a_1 = a_2 = \cdots = a_{n-1} = 1\) and \(a_n = n + 1\).

Conclusion:
The *good* \(n\)-tuples are:
1. \((2, 2, \ldots, 2)\)
2. \((1, 1, \ldots, 1, n+1)\)

The final answer is \( \boxed{ (2, 2, \ldots, 2) } \) and \((1, 1, \ldots, 1, n+1)\)