$2 \cdot 31$ Find the smallest positive integer $n$, such that among any $n$ irrational numbers, there always exist 3 numbers, the sum of any two of which is still irrational.
[Solution] Clearly, any 3 numbers among the 4 numbers $\{\sqrt{2}, \sqrt{2},-\sqrt{2},-\sqrt{2}\}$ include a pair of opposite numbers, whose sum is 0, which is of course not an irrational number. Therefore, the smallest positive integer $n \geqslant 5$.

Let $x, y, z, u, v$ be 5 irrational numbers. We use 5 points to represent them. If the sum of two numbers is a rational number, then we draw a red line between the corresponding two points; otherwise, we draw a blue line. Thus, we obtain a 2-colored complete graph $K_{5}$ with 5 vertices. We need to prove that there must be a blue triangle in the graph.

If not, then the graph either has a red triangle or has no monochromatic triangle (a triangle with all 3 sides the same color).

If the former is the case, let $\triangle X Y Z$ be a red triangle. Then the 3 sums $x+y, y+z, z+x$ are all rational numbers. Therefore,
$$
x=\frac{1}{2}\{(x+y)+(x+z)-(y+z)\}
$$

is also a rational number, which is a contradiction.
If the latter is the case, i.e., the 2-colored $K_{5}$ has no monochromatic triangle. It is easy to prove that in this case, each point has 4 edges, 2 red and 2 blue, and the 5 red edges and 5 blue edges each form a cycle with 5 edges. Without loss of generality, let the red cycle be $X Y Z U V$. Then the 5 sums $x+y, y+z, z+u$, $u+v, v+x$ are all rational numbers. Therefore,
$$
x=\frac{1}{2}\{(x+y)+(z+u)+(v+x)-(y+z)-(u+v)\}
$$

is also a rational number, which is a contradiction.