Since no one else does i'll post a solution. By means of inequalities one can show that $n\le7$. The possible values of $n$ are $3,4,5,6,7$. The corresponding $2n$-tuples are n=3: $(a_1,a_2,a_3,a_4,a_5,a_6)=(1,2,3,6,4,5)$ n=4:$(2,3,1,6,7,4,8,5)$ n=5:$(2,4,1,6,3,7,9,5,8,10)$ n=6:$(5,8,4,9,3,7,2,6,1,11,10,12)$ n=7:$(1,12,2,11,3,10,4,9,5,8,6,7,13,14)$

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bigant146 wrote: Find all positive integers $n\geqslant 2$ such that there exists a permutation $a_1$, $a_2$, $a_3$, \ldots, $a_{2n}$ of the numbers $1, 2, 3, \ldots, 2n$ satisfying $$a_1\cdot a_2 + a_3\cdot a_4 + \ldots + a_{2n-3} \cdot a_{2n-2} = a_{2n-1} \cdot a_{2n}.$$ $$a_1\cdot a_2 + a_3\cdot a_4 + \ldots + a_{2n-3} \cdot a_{2n-2} + a_{2n-1} \cdot a_{2n}= 2a_{2n-1} \cdot a_{2n}.$$$(a_1,a_2,...,a_{2n})=(1,2,...,2n)$ $$2A= 2(a_1\cdot a_2 + a_3\cdot a_4 + \ldots + a_{2n-3} \cdot a_{2n-2} + a_{2n-1} \cdot a_{2n})$$$$B=(a_1^2+a_2^2)+...+(a_{2n-1}^2+a_{2n}^2)$$$$B+2A=(a_1+a_2)^2+...+(a_{2n-1}+a_{2n})^2$$$$[(a_1+a_2)^2+...+(a_{2n-1}+a_{2n})^2]\cdot[1+1+...+1] \geq (a_1+a_2 + ... + a_{2n-1}+a_{2n})^2 = (n(2n+1))^2$$$$[B+2A]n \geq n^2(2n+1)^2$$$$B+2A \geq n(2n+1)^2$$$$B = 1^2+2^2+...+(2n)^2 = \frac{2n(2n+1)(4n+1)}{6}$$$$2(2n)(2n-1) \geq 2a_{2n-1}a_{2n}=A\geq \frac{n(n+1)(2n+1)}{3}$$Remaining easy.$\blacksquare$