A natural number $n$ is called [i]perfect [/i] if it is equal to the sum of all its natural divisors other than $n$. For example, the number $6$ is perfect because $6 = 1 + 2 + 3$. Find all even perfect numbers that can be given as the sum of two cubes positive integers.

1. **Understanding the problem and the theorem**:
   - A natural number \( n \) is called *perfect* if it is equal to the sum of all its natural divisors other than \( n \). For example, \( 6 \) is perfect because \( 6 = 1 + 2 + 3 \).
   - By a theorem of Euler-Euclid, an even perfect number \( n \) can be expressed as \( n = 2^{p-1}(2^p - 1) \), where \( 2^p - 1 \) is a Mersenne prime, and \( p \) is a prime number.

2. **Checking small values of \( p \)**:
   - For \( p = 2 \), \( n = 2^{2-1}(2^2 - 1) = 2 \cdot 3 = 6 \). However, \( 6 \) cannot be expressed as the sum of two cubes of positive integers.
   - For \( p = 3 \), \( n = 2^{3-1}(2^3 - 1) = 4 \cdot 7 = 28 \). We need to check if \( 28 \) can be expressed as the sum of two cubes of positive integers.

3. **Checking \( n = 28 \)**:
   - We need to find \( a \) and \( b \) such that \( 28 = a^3 + b^3 \).
   - Trying small values, we find that \( 28 \) cannot be expressed as the sum of two cubes of positive integers.

4. **General case for \( p \geq 5 \)**:
   - Suppose \( p \geq 5 \). We need to solve \( 2^{p-1}(2^p - 1) = a^3 + b^3 \).
   - Note that \( a \) and \( b \) must have the same parity (both odd or both even).

5. **Case 1: \( a \) and \( b \) are both odd**:
   - If \( a \) and \( b \) are both odd, then \( 2^{p-1}(2^p - 1) = (a + b)(a^2 - ab + b^2) \).
   - Since \( a^2 - ab + b^2 \) is odd, we have \( a^2 - ab + b^2 = 2^p - 1 \) and \( a + b = 2^{p-1} \).
   - Using the inequality \( a^2 - ab + b^2 \geq ab \), we get \( (a - 2)(b - 2) \leq 3 \).
   - Solving this, we find that \( a = 3 \) and \( b = 1 \) is the only solution, yielding \( n = 28 = 2^2(2^3 - 1) \).

6. **Case 2: \( a \) and \( b \) are both even**:
   - Let \( a + b = 2^\theta \) and \( a^2 - ab + b^2 = 2^{p-1-\theta}(2^p - 1) \).
   - Let \( b = 2^\theta - a \). Then, \( 3a^2 - 3 \cdot 2^\theta a + 2^{2\theta} = (2^p - 1)2^{p-1-\theta} \).
   - Let \( v_2(a) = i \). Since \( a < 2^\theta \), clearly \( i < \theta \).
   - Thus, \( v_2(3a^2 - 3 \cdot 2^\theta a + 2^{2\theta}) = 2i = p-1-\theta \).
   - Hence, \( 2^{\frac{p-1-\theta}{2}} \mid\mid a \).
   - Using \( (a + b)^2 - (a^2 - ab + b^2) = 3ab \), we conclude that \( 3ab = 2^{2\theta} - (2^p - 1)2^{p-1-\theta} \).
   - Consequently, \( v_2(a + b) = p-1-\theta \). Thus, \( 2^{\frac{p-1-\theta}{2}} \mid b \).
   - Let \( a = 2^{\frac{p-1-\theta}{2}} a_1 \) and \( b = 2^{\frac{p-1-\theta}{2}} b_1 \).
   - With this, we obtain \( a_1^2 - a_1b_1 + b_1^2 = 2^p - 1 \).
   - Moreover, \( 2^{\frac{3\theta-(p-1)}{2}} = a_1 + b_1 \).
   - Thus, \( a_1 \) and \( b_1 \) have the same parity. If both are even, \( a_1^2 - a_1b_1 + b_1^2 \) is even, which is not possible.
   - Thus, both \( a_1 \) and \( b_1 \) are odd. If \( a_1 + b_1 > 6 \), then \( a_1 + b_1 \equiv 0 \pmod{8} \).
   - With this, \( b_1 \equiv -a_1 \pmod{8} \), hence \( a_1^2 - a_1b_1 + b_1^2 \equiv 3a_1^2 \equiv 3 \pmod{8} \), whereas \( 2^p - 1 \equiv -1 \pmod{8} \). This is a contradiction.
   - Hence, \( a_1 + b_1 \in \{2, 4, 6\} \) with \( a_1, b_1 \) being odd.
   - If \( a_1 + b_1 = 2 \), then \( a_1^2 - a_1b_1 + b_1^2 = 1 \), not a Mersenne prime.
   - If \( a_1 + b_1 = 4 \) with \( a_1, b_1 \) odd, then the only possibility is \( a_1 = 3, b_1 = 1 \) (or vice versa) in which case we have \( a_1^2 - a_1b_1 + b_1^2 = 7 \), that is \( p = 3 \).
   - Finally, for \( a_1 + b_1 = 6 \), we have either \( a_1 = 1, b_1 = 5 \) (or vice versa), or \( a_1 = b_1 = 3 \). In the former, \( a_1^2 - a_1b_1 + b_1^2 = 21 \), not a Mersenne prime. In the latter, \( 3 \mid a_1^2 - a_1b_1 + b_1^2 \), not possible.

7. **Conclusion**:
   - Thus, \( n = 28 \) is indeed the only even perfect number that can be expressed as the sum of two cubes of positive integers.

The final answer is \( \boxed{ 28 } \).