Solution:
Let $ABC$ be a triangle with $BC=a$, $AC=b$, $AB=c$ and $a<b<c$. Denote by $m_{a}, m_{b}, m_{c}$ the lengths of medians drawn from the vertices $A, B, C$ respectively. Using the formulas
$$4 m_{a}^{2} = 2(b^{2} + c^{2}) - a^{2}, \quad 4 m_{b}^{2} = 2(a^{2} + c^{2}) - b^{2}, \quad 4 m_{c}^{2} = 2(a^{2} + b^{2}) - c^{2}$$
we obtain the relations
$$\begin{gathered} 4 m_{a}^{2} + 4 m_{b}^{2} + 4 m_{c}^{2} = 3a^{2} + 3b^{2} + 3c^{2} \\ \left(4 m_{a}^{2}\right)^{2} + \left(4 m_{b}^{2}\right)^{2} + \left(4 m_{c}^{2}\right)^{2} = \left(2b^{2} + 2c^{2} - a^{2}\right)^{2} + \\ \left(2a^{2} + 2c^{2} - b^{2}\right)^{2} + \left(2a^{2} + 2b^{2} - c^{2}\right)^{2} = 9a^{4} + 9b^{4} + 9c^{4} = \\ \left(3a^{2}\right)^{2} + \left(3b^{2}\right)^{2} + \left(3c^{2}\right)^{2} \end{gathered}$$
We put $A=\{4 m_{a}^{2}, 4 m_{b}^{2}, 4 m_{c}^{2}\}$ and $B=\{3a^{2}, 3b^{2}, 3c^{2}\}$. Let $k \geq 1$ be a positive integer. Let $a = k+1$, $b = k+2$ and $c = k+3$. Because
$$a+b = (k+1) + (k+2) = 2k+3 > k+3 = c$$
a triangle with such side lengths exists. After simple calculations we have
$$\begin{gathered} A = \{3(k+1)^{2} - 2,\ 3(k+2)^{2} + 4,\ 3(k+3)^{2} - 2\} \\ B = \{3(k+1)^{2},\ 3(k+2)^{2},\ 3(k+3)^{2}\} \end{gathered}$$
It is easy to prove that
$$\begin{gathered} x + y + z = m + n + p = 3\left[(k+1)^{2} + (k+2)^{2} + (k+3)^{2}\right] \\ x^{2} + y^{2} + z^{2} = m^{2} + n^{2} + p^{2} = 9\left[(k+1)^{4} + (k+2)^{4} + (k+3)^{4}\right] \end{gathered}$$
From the inequality $3(k+1)^{2} - 2 > 2003$ we obtain $k \geq 25$. For $k=25$ we have an example of two sets
$$A = \{2026, 2191, 2350\}, \quad B = \{2028, 2187, 2352\}$$
with the desired properties.