2. [4] Find all natural numbers n that satisfy the following condition: the numbers 1, 2, 3, ..., 2n can be paired in such a way that if the numbers in each pair are added and the results are multiplied, the product is a square of a natural number.

(Folklore)
Answer: all $n>1$. Solution. $1+2$ is not a square. Let $n>1$.

1st method. We will divide these numbers into groups of four consecutive numbers, and, if necessary, a group of six initial numbers. From the groups of four, we form $(a+(a+3))((a+1)+(a+2))=(2 a+3)^{2}$, and from the group of six - $(1+5)(2+4)(3+6)=18^{2}$.

2nd method. If $n$ is even, then $(1+2 n)(2+(2 n-1)) \ldots(n+(n+1))=(2 n+1)^{n}$ - a square. If $n$ is odd, then $(1+5)(2+4)(3+6)(7+2 n)(8+(2 n-1)) \ldots((n+3)+(n+4))=18^{2}(2 n+7)^{n-3}$ - a square.

Remark. For $n=2,3$, the partition is unique; in other cases, it is not.