2. Let natural numbers $x, y$ satisfy
$$
x<y, x^{3}+19 y=y^{3}+19 x
$$

Then $x+y=$ $\qquad$
2.5.

From $x^{3}-y^{3}=19(x-y), x<y$, we get $x^{2}+x y+y^{2}=19$.
Thus, $3 x^{2}<x^{2}+x y+y^{2}=19$.
Since $x$ is a natural number, we have $x=2$.
Therefore, $y=3$, so $x+y=5$.