USAMO

This problem is a form of Chebyshev's inequality for random variables. For each subset $A \subseteq \{1, 2, \dots, n\}$, define
$$\Delta_A = 2S_A = \sum_{i \in A} x_i - \sum_{i \in \{1, 2, \dots, n\} \setminus A} x_i = \sum_{i=1}^n \epsilon_A(i)x_i,$$
where $\epsilon_A(i) = 1$ if $i \in A$ and $\epsilon_A(i) = -1$ otherwise. Squaring, we have
$$\Delta_A^2 = \sum_{i=1}^n x_i^2 + \sum_{\substack{i,j \in \{1, \dots, n\} \\ i \neq j}} \epsilon_A(i)\epsilon_A(j)x_i x_j. \quad (10)$$
Now sum the $\Delta_A^2$'s over all $2^n$ possible choices of $A$. For each pair $i \neq j$, there are $2^{n-2}$ sets $A$ with $i, j \in A$, and another $2^{n-2}$ sets with $i, j \notin A$; these sets each contribute a term of $+x_i x_j$ to the sum in (10). There are also $2^{n-2}$ sets $A$ with $i \in A$, $j \notin A$, and $2^{n-2}$ sets with $i \notin A$, $j \in A$. Each of these sets contributes a term of $-x_i x_j$ to (10). Hence $x_i x_j$ appears $2^{n-1}$ times with a $+$ sign and $2^{n-1}$ times with a $-$ sign. Therefore all of these terms cancel, and we obtain
$$\sum_{A \subseteq \{1, 2, \dots, n\}} \Delta_A^2 = 2^n (x_1^2 + \dots + x_n^2) = 2^n. \quad (11)$$
Now let $\lambda > 0$. There cannot be more than $2^{n-2}/\lambda^2$ terms $\Delta_A^2$ whose value greater than or equal to $4\lambda^2$. If this were not the case, then the sum of these terms would be greater than $2^n$, so the sum in (11) would exceed $2^n$. Hence, there can be at most $2^{n-2}/\lambda^2$ sets $A$ such that $|S_A| \ge \lambda$. (Recall that $\Delta_A = 2S_A$.) Moreover, these sets can be arranged into complementary pairs because $S_A = -S_{\{1, \dots, n\} \setminus A}$. In each of these pairs, exactly one of the two members is positive. Therefore there are at most $2^{n-3}/\lambda^2$ sets $A$ with $S_A \ge \lambda$.

For equality to hold, it must be the case that all positive values of $\Delta_A^2$ are equal to $4\lambda^2$; otherwise we would again have a contradiction because the sum of all $\Delta_A^2$ would exceed $2^n$. In particular, all positive values of $\Delta_A^2$ must be the same. Thus, all positive values of $x_A$ must be the same. This will be the case only if at most one of the $x_i$ is positive and at most one of the $x_i$ is negative. Because we must have at least one of each, there must be exactly one positive term and one negative term. Thus, it must be the case that $x_k = \sqrt{2}/2$ for some $k$, $x_j = -\sqrt{2}/2$ for some $j \neq k$, and $x_i = 0$ for $i \neq j, k$. Then the assumption that every positive $\Delta_A^2 = 4\lambda^2$ yields $\lambda = \sqrt{2}/2$.

Conversely, with the $x_i$ and $\lambda$ as described, we have exactly $2^{n-2} = 2^{n-3}/\lambda^2$ sets $A$ such that $x_A \ge \lambda$: namely, the sets $A$ that contain the $\sqrt{2}/2$ term and do not contain the $-\sqrt{2}/2$ term. Thus, this is indeed the equality case.