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 $1^2 + 2^2 + \dots + p^2 = \frac{p(p+1)(2p+1)}{6}$. So for $p > 3$, $p$ divides $1^2 + 2^2 + \dots + p^2$, 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.

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

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 $n \equiv -1 \pmod 3$ or $n \equiv 1 \pmod 3$. Therefore splitting the numbers $1, 2, 3, \dots, 2n$ into the pairs $(1, 2), (3, 4), \dots, (2n-1, 2n)$ works for infinitely many $n$.

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 $p \equiv 1 \pmod 4$ (by Dirichlet's theorem, there are infinitely many such primes), then there exists an integer $a$ such that $a^2 \equiv -1 \pmod p$. If $p > 2$, then clearly $a \not\equiv 1, -1 \pmod p$. Then for all $i = 1, 2, \dots, p-1$ the numbers $i, ai, a^2i, a^3i$ give different remainders modulo $p$, whereas $a^4i \equiv i \pmod p$. 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 $i^2 + (ai)^2 = (1 + a^2)i^2 \equiv (1 - 1)i^2 = 0 \pmod p$. So the sum $1^2 + 2^2 + \dots + (p-1)^2$ 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.