Estonian Math Competitions

For each prime $p$ we can split the numbers $1,2,3,\dots ,2p$ into the pairs $(1,p+1),(2,p+2),\dots ,(p,2p)$. The products of the pairs are congruent to $1^2,2^2,\dots ,(p-1{)}^2,p^2$ modulo $p$, so the sum of the products is congruent to $1^2+2^2+\cdots +p^2=\frac{p(p+1)(2p+1)}{6}$. So for $p>3$, $p$ divides $1^2+2^2+\cdots +p^2$, as it doesn't divide the denominator. Also, in each pair one of the numbers is even, so the product is even. Hence the sum of the products is divisible by $2$ and therefore also by $2p$, as desired.

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Alternative solution.

We use the identity $1\cdot 2+3\cdot 4+\cdots +(2n-1)\cdot 2n=\frac{n(n+1)(4n-1)}{3}$.

If $n$ is not divisible by $2$ or $3$, then $2n\mid \frac{n(n+1)(4n-1)}{3}$, since $2\mid n+1$ and either $3\mid n+1$ or $3\mid 4n-1$, depending on whether $n\equiv -1(\mathrm{mod}3)$ or $n\equiv 1(\mathrm{mod}3)$. Therefore splitting the numbers $1,2,3,\dots ,2n$ into the pairs $(1,2),(3,4),\dots ,(2n-1,2n)$ works for infinitely many $n$.

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Alternative solution.

For each prime $p$ we can split the numbers $1,2,3,\dots ,2p$ into the pairs $(1,p+1),(2,p+2),\dots ,(p,2p)$. The products of the pairs are congruent to $1^2,2^2,\dots ,(p-1{)}^2,p^2$ modulo $p$. Let $p\equiv 1(\mathrm{mod}4)$ (by Dirichlet's theorem, there are infinitely many such primes), then there exists an integer $a$ such that $a^2\equiv -1(\mathrm{mod}p)$. If $p>2$, then clearly $a\not\equiv 1,-1(\mathrm{mod}p)$. Then for all $i=1,2,\dots ,p-1$ the numbers $i,ai,a^{2}i,a^{3}i$ give different remainders modulo $p$, whereas $a^{4}i\equiv i(\mathrm{mod}p)$. So the remainders $1,2,\dots ,p-1$ are split into 4-cycles. The sum of squares of each 4-cycle is divisible by $p$, as $i^2+(ai{)}^2=(1+a^2)i^2\equiv (1-1)i^2=0(\mathrm{mod}p)$. So the sum $1^2+2^2+\cdots +(p-1{)}^2$ is divisible by $p$. Adding also the final term $p^2$, the divisibility will still hold. Also, in each pair one of the numbers is even, so the product is even. Therefore the sum of the products is divisible by $2$ and therefore also by $2p$, as desired.