Given a natural number $n$, if the tuple $(x_1,x_2,\ldots,x_k)$ satisfies
$$2\mid x_1,x_2,\ldots,x_k$$
$$x_1+x_2+\ldots+x_k=n$$
then we say that it's an [i]even partition[/i]. We define [i]odd partition[/i] in a similar way. Determine all $n$ such that the number of even partitions is equal to the number of odd partitions.
1. **Understanding the Problem:**
   We need to determine all natural numbers \( n \) such that the number of even partitions of \( n \) is equal to the number of odd partitions of \( n \). An even partition is a partition where all parts are even, and an odd partition is a partition where all parts are odd.

2. **Case Analysis for Odd \( n \):**
   If \( n \) is odd, then there cannot be any even partitions because the sum of even numbers is always even. Therefore, the number of even partitions is zero. However, there is at least one odd partition (e.g., \( (n) \) itself). Hence, for odd \( n \), the number of even partitions is not equal to the number of odd partitions.

3. **Case Analysis for Even \( n \):**
   Let \( n \) be even. We need to compare the number of even partitions and odd partitions.

4. **Constructing Odd Partitions from Even Partitions:**
   Given an even partition \( (x_1, x_2, \ldots, x_k) \) where \( x_i \) are all even, we can construct an odd partition by subtracting 1 from each \( x_i \) and adding \( k \) ones:
   \[
   (x_1 - 1, x_2 - 1, \ldots, x_k - 1, 1, 1, \ldots, 1)
   \]
   This transformation is injective because we can recover the original even partition by reversing the process: count the number of 1's to determine \( k \), then add 1 to each of the remaining numbers.

5. **Checking for Missing Partitions:**
   For \( n \geq 6 \), consider the odd partition \( (n-3, 3) \). This partition cannot be obtained from any even partition using the above transformation because it would require an even partition of \( n-2 \), which is not possible since \( n-2 \) is even and cannot be partitioned into all even parts.

6. **Special Cases \( n = 2 \) and \( n = 4 \):**
   - For \( n = 2 \):
     - Even partitions: \( (2) \)
     - Odd partitions: \( (1, 1) \)
     - Both counts are 1.
   - For \( n = 4 \):
     - Even partitions: \( (4), (2, 2) \)
     - Odd partitions: \( (3, 1), (1, 1, 1, 1) \)
     - Both counts are 2.

7. **Conclusion:**
   For \( n = 2 \) and \( n = 4 \), the number of even partitions is equal to the number of odd partitions. For \( n \geq 6 \), the number of odd partitions is greater than the number of even partitions.

The final answer is \( \boxed{ n = 2, 4 } \)