For which positive integers $n$ is the following statement true: 
if $a_1, a_2, ... , a_n$ are positive integers, $a_k \le n$ for all $k$ and $\sum\limits_{k=1}^{{n}}{a_k}=2n$ 
then it is always possible to choose $a_{i1} , a_{i2} , ..., a_{ij}$ in such a way that 
the indices $i_1, i_2,... , i_j$ are different numbers, and  $\sum\limits_{k=1}^{{{j}}}{a_{ik}}=n$?
1. **Initial Considerations:**
   - We need to determine for which positive integers \( n \) the given statement holds.
   - The statement involves a sequence of positive integers \( a_1, a_2, \ldots, a_n \) such that \( a_k \le n \) for all \( k \) and \(\sum_{k=1}^{n} a_k = 2n\).
   - We need to find a subset of these integers whose sum is exactly \( n \).

2. **Case Analysis for Small Values of \( n \):**
   - For \( n = 1 \):
     \[
     \sum_{k=1}^{1} a_k = a_1 = 2 \quad \text{(impossible since } a_1 \le 1 \text{)}
     \]
     Thus, \( n = 1 \) is not possible.

   - For \( n = 2 \):
     \[
     a_1 + a_2 = 4 \quad \text{and} \quad a_1, a_2 \le 2
     \]
     The only possible values are \( a_1 = a_2 = 2 \). We can choose \( a_1 \) or \( a_2 \) to get a sum of \( 2 \). Thus, \( n = 2 \) works.

3. **General Case for Even \( n \):**
   - Assume \( n \) is even. We need to show that for any sequence \( a_1, a_2, \ldots, a_n \) satisfying the conditions, we can find a subset summing to \( n \).

4. **Sequence Analysis:**
   - Assume \( a_1 \le a_2 \le \ldots \le a_n \).
   - If \( a_1 = a_2 = \ldots = a_n = 2 \), then we can choose any \( n/2 \) of these to get a sum of \( n \).

5. **General Sequence with \( a_1 = 1 \):**
   - Suppose there are \( x \) ones and \( y \) twos in the sequence. The remaining \( n - x - y \) numbers are at least three.
   - We have:
     \[
     x + 2y + 3(n - x - y) \le 2n
     \]
     Simplifying, we get:
     \[
     3n - 2x - y \le 2n \implies n \le 2x + y
     \]

6. **Subset Sum Analysis:**
   - If \( 2x + y \ge n \), we can choose a subset from the first \( x + y \) elements to get a sum of \( n \).
   - If \( 2x + y < n \), consider the remaining elements \( a_{x+y+1}, a_{x+y+2}, \ldots, a_n \). If any of these is at least \( n/2 \), we can pair it with elements from the first \( x + y \) to get a sum of \( n \).

7. **Maximal Subset Sum:**
   - Let \( \{a_{k_1}, a_{k_2}, \ldots, a_{k_l}\} \) be the subset of \( \{a_{x+y+1}, a_{x+y+2}, \ldots, a_n\} \) with the largest sum less than \( n/2 \).
   - There must exist some \( a_i \) in \( \{a_{x+y+1}, a_{x+y+2}, \ldots, a_n\} \setminus \{a_{k_1}, a_{k_2}, \ldots, a_{k_l}\} \) such that adding \( a_i \) to the sum of \( \{a_{k_1}, a_{k_2}, \ldots, a_{k_l}\} \) gives a sum between \( n/2 \) and \( n \).

8. **Conclusion:**
   - We can always find a subset of the sequence summing to \( n \) if \( n \) is even.

The final answer is \( \boxed{ n } \) is even.