Solution: Consider the four numbers $\sqrt{2}, \sqrt{2}, -\sqrt{2}, -\sqrt{2}$, we can see that $n \geqslant 5$.

When $n=5$, let the 5 numbers be $x, y, z, u, v$. Corresponding to 5 points $X, Y, Z, U, V$.

Connect two points with a blue edge if and only if the sum of the two corresponding irrational numbers is irrational; otherwise, connect the two points with a red edge, forming graph $G$.

We will prove: (1) There is no 5-sided red cycle in the graph; (2) There is no red triangle in the graph.

First, prove (2). If there is a red triangle in the graph, without loss of generality, let it be triangle $XYZ$, i.e., $x+y, y+z, z+x$ are all rational numbers. Then $2y = (x+y) + (y+z) - (x+z)$ is a rational number, which is a contradiction.
Next, prove (1). If there is a 5-sided red cycle $XYZUV$ in the graph, i.e., $x+y, y+z, z+u, u+v, v+x$ are all rational numbers, then
$$
2x = (x+y) + (z+u) + (z+x) - (y+z) - (u+v)
$$
is a rational number, which is a contradiction!
Thus, (1) and (2) are proven. By Theorem 1 and Theorem 2, the graph contains a blue triangle.
Therefore, $n_{\min} = 5$.
3. Find the smallest natural number $n$, such that among any $n$ irrational numbers, there always exist 3 numbers, the sum of any two of which is still an irrational number.