Hkg TST

The average score is $670000$.
Let $n = 100$, and let $S$ be any subset of $\{1, 2, \dots, 2n\}$ such that $|S| = n$. There are $\binom{2n}{n}$ such sets $S$. The score of $S$ is
$$\begin{aligned} \left( \sum_{a \in S} a - \sum_{b \notin S} b \right)^2 &= \left( \sum_{k=1}^{2n} k - 2 \sum_{a \in S} a \right)^2 \\ &= \left( \frac{2n(2n+1)}{2} \right)^2 - 4 \left( \frac{2n(2n+1)}{2} \right) \sum_{a \in S} a + 4 \left( \sum_{a \in S} a \right)^2 \\ &= n^2(2n+1)^2 - 4n(2n+1) \sum_{a \in S} a + 4 \sum_{a \in S} a^2 + 4 \sum_{\substack{a, b \in S \\ a \neq b}} ab. \end{aligned}$$

$$\sum_{S} \sum_{a \in S} a = \sum_{a=1}^{2n} \sum_{a \in S} a = \sum_{a=1}^{2n} a \binom{2n-1}{n-1} = n(2n+1) \binom{2n-1}{n-1}$$

$$\begin{align*} \sum_S \left( \sum_{a \in S} a^2 + \sum_{\substack{a, b \in S \\ a \neq b}} ab \right) &= \sum_{a=1}^{2n} \sum_{a \in S} a^2 + \sum_{\substack{a, b=1 \\ a \neq b}}^{2n} \sum_{a, b \in S} ab \\ &= \sum_{a=1}^{2n} a^2 \binom{2n-1}{n-1} + \sum_{\substack{a, b=1 \\ a \neq b}}^{2n} ab \binom{2n-2}{n-2} \\ &= \sum_{a=1}^{2n} a^2 \binom{2n-1}{n-1} + \sum_{a=1}^{2n} a \left( \sum_{b=1}^{2n} b - a \right) \binom{2n-2}{n-2} \\ &= \sum_{a=1}^{2n} a^2 \left[ \binom{2n-1}{n-1} - \binom{2n-2}{n-2} \right] + n(2n+1) \binom{2n-2}{n-2} \sum_{a=1}^{2n} a \\ &= \frac{n(2n+1)(4n+1)}{3} \binom{2n-2}{n-1} + n^2(2n+1)^2 \binom{2n-2}{n-2}. \end{align*}$$

Therefore, the average score is
$$\begin{align*} & n^2(2n+1)^2 + \left(\frac{2n}{n}\right)^{-1} \left[ -4n^2(2n+1)^2 \binom{2n-1}{n-1} + \frac{4n(2n+1)(4n+1)}{3} \binom{2n-2}{n-1} \right. \\ & \qquad \left. + 4n^2(2n+1)^2 \binom{2n-2}{n-2} \right] \\ &= n^2(2n+1)^2 + \left( -2n^2(2n+1)^2 + \frac{2n^2(2n+1)(4n+1)}{3(2n-1)} + \frac{2(n-1)n^2(2n+1)^2}{2n-1} \right) \\ &= \frac{n^2(2n+1)}{3} \\ &= 670000. \end{align*}$$