Iranian Mathematical Olympiad

Let's place these $2n-1$ numbers around a circle and label them by $\pi(1), \pi(2), \dots, \pi(2n-1)$ respectively and let
$$S_i = \pi(i) + \pi(i+1) + \dots + \pi(i + n - 1), \quad i \le 2n-1$$
(Numbers are considered modulo $2n-1$.) We claim that at least $n$ of $S_i$'s have a value not less than $\frac{S}{2}$.

Assume the contrary which means at least $n$ of these $S_i$'s have a value less than $\frac{S}{2}$. Let the indices of these $S_i$'s be $i_1, i_2, \dots, i_n$. Note that, by the pigeonhole principle there are two indices $i_s, i_t$ such that
$$i_t - i_s \stackrel{2n-1}{\equiv} n$$
(Because there is a total of $n-1$ sets of the form of $\{i, i+n\}$ (mod $2n-1$), and we have chosen $n$ indices, each one exists in a set.) So we can get
$$\begin{align*} \frac{S}{2} \times 2 > S_{i_s} + S_{i_t} &= \pi(i_s) + \pi(i_s + 1) + \dots + \pi(i_s + n - 1) \\ &\quad + \pi(i_t) + \pi(i_t + 1) + \dots + \pi(i_t + n - 1) \\ &= \pi(1) + \pi(2) + \dots + \pi(2n - 1) + \pi(i_s) > S \end{align*}$$
Which is a contradiction. Hence the claim.

Now there is a total of $(2n-2)!$ ways to place these numbers around the circle. Also according to the claim, in each permutation we have at least $n$ sums satisfying the condition. Each sum exists in $n! \times (n-1)!$ permutations. So there's a total of at least
$$\frac{(2n-2)! \times n}{n! \times (n-1)!} = \binom{2n-2}{n-1}$$
ways to choose an $n$-tuple with sum greater or equal to $\frac{S}{2}$.