Are there such pairwise distinct natural numbers $ m, n, p, q$ satisfying $ m \plus{} n \equal{} p \plus{} q$ and $ \sqrt{m} \plus{} \sqrt[3]{n} \equal{} \sqrt{p} \plus{} \sqrt[3]{q} > 2004$ ?
1. We start with the given conditions:
   \[
   m + n = p + q \quad \text{and} \quad \sqrt{m} + \sqrt[3]{n} = \sqrt{p} + \sqrt[3]{q} > 2004
   \]

2. Let \( n = b^3 \) and \( q = a^3 \). Then we can express \( m \) and \( p \) as:
   \[
   m = k - b^3 \quad \text{and} \quad p = k - a^3
   \]
   where \( k = m + n = p + q \).

3. We need to find a solution to:
   \[
   \sqrt{k - b^3} + b = \sqrt{k - a^3} + a
   \]

4. Let \( k = b^3 + y^2 \). Then \( m = y^2 \). We need to solve:
   \[
   y + b = \sqrt{b^3 + y^2 - a^3} + a
   \]

5. Squaring both sides:
   \[
   (y + b - a)^2 = b^3 + y^2 - a^3
   \]

6. Expanding and simplifying:
   \[
   y^2 + 2yb + b^2 - 2ya - 2ab + a^2 = b^3 + y^2 - a^3
   \]
   \[
   2yb + b^2 - 2ya - 2ab + a^2 = b^3 - a^3
   \]

7. Factoring out common terms:
   \[
   (b - a)(b - a + 2y) = (b - a)(b^2 + ab + a^2)
   \]

8. Since \( b \neq a \), we can divide both sides by \( b - a \):
   \[
   b - a + 2y = b^2 + ab + a^2
   \]

9. Solving for \( y \):
   \[
   2y = b^2 + ab + a^2 + a - b
   \]
   \[
   y = \frac{b^2 + ab + a^2 + a - b}{2}
   \]

10. Let \( a = 2r \) and \( b = 2 \). Then:
    \[
    y = \frac{2^2 + 2 \cdot 2r + (2r)^2 + 2r - 2}{2}
    \]
    \[
    y = \frac{4 + 4r + 4r^2 + 2r - 2}{2}
    \]
    \[
    y = \frac{4r^2 + 6r + 2}{2}
    \]
    \[
    y = 2r^2 + 3r + 1
    \]

11. Therefore:
    \[
    n = 8, \quad q = 8r^3
    \]
    \[
    m = (2r^2 + 3r + 1)^2 = 4r^4 + 12r^3 + 13r^2 + 6r + 1
    \]
    \[
    p = (2r^2 + r + 3)^2 = 4r^4 + 4r^3 + 13r^2 + 6r + 9
    \]

12. We can make \( r \) arbitrarily large to satisfy \( \sqrt{m} + \sqrt[3]{n} = \sqrt{p} + \sqrt[3]{q} > 2004 \).

The final answer is \( \boxed{ m, n, p, q } \) satisfying the given conditions.