Baltic Way competition problems

Answer:
$$\frac{(n-2)(2n^2 - 5n)}{24}$$

Solution:
Letting $a$ be the smallest and $b$ the largest of the chosen numbers, the sum in the problem has to be $a+b$. So given $a$ and $b$, the other numbers $c$ and $d$ have to satisfy $a < c, d < b$ and $c+d = a+b$. For the smaller of $c, d$, say $c$, one can take any number larger than $a$ but smaller than the average of $a$ and $b$, and the choice $c$ uniquely determines $d$. So if $b-a = 2p+1$ or $b-a = 2p+2$, there are $p$ possible choices of $c$. Assume $n=2m$. Then the largest possible $b-a = 2m-1 = 2(m-1)+1$, and there is just one possible pair $(a, b) = (1, 2m) = (1, n)$. For $b-a = n-q$ there are $q$ possible pairs $(a, b)$. The $p$ possible choices of $c$ thus appear when $q = n-2p-2$ and $q = n-2p-1$, or altogether in $2n-4p-3$ cases. So the total number of choices is
$$\begin{aligned} \sum_{p=1}^{m-2} (2n - 4p - 3)p + m - 1 &= (2n - 3) \sum_{p=1}^{m-2} p - 4 \sum_{p=1}^{m-2} p^2 + m - 1 \\ &= \frac{1}{2}(2n - 3)(m - 2)(m - 1) - \frac{4}{6}(m - 2)(m - 1)(2m - 3) + (m - 1). \end{aligned}$$
Using $2m=n$, the last sum is easily simplified into
$$\frac{(n-2)(2n^2 - 5n)}{24}$$
The restriction "n even" can be removed, but then there are two essentially similar but slightly different sums to be done. "n odd" would be infinitesimally easier, because there would not be the single last term.