Problem 1762

Let $M(n)$ be the maximum of $\sum_{(i,j) \in T} x_i x_j$. We will show that
$$M(n) = \frac{\lfloor \log_2 n \rfloor}{2(\lfloor \log_2 n \rfloor + 1)}.$$
For $k = \lfloor \log_2 n \rfloor$, set $x_{20} = x_{21} = \dots = x_{2k} = \frac{1}{k+1}$, and $x_i = 0$ otherwise. Then we have $\sum_{(i,j) \in T} x_i x_j = \binom{k+1}{2} \frac{1}{(k+1)^2} = \frac{k}{2(k+1)}$, thus
$$M(n) \ge \frac{\lfloor \log_2 n \rfloor}{2(\lfloor \log_2 n \rfloor + 1)}.$$
Consider an element $(x_1, x_2, \dots, x_n)$ which yields $M(n)$ and the number of $i$'s such that $x_i = 0$ is maximal. In this case, if $x_a, x_b \ne 0$ and $a < b$, then we have $(a, b) \in T$. Suppose $(a, b) \notin T$, then by setting $x'_a = x_a - \epsilon$, $x'_b = x_b + \epsilon$, and $x'_i = x_i$ ($i \ne a, b$), the equation $x'_1 + x'_2 + \dots + x'_n = 1$ still holds and the value $\sum_{(i,j) \in T} x'_i x'_j$ becomes a linear function of $\epsilon$. Thus for $\epsilon = x'_a$ or $\epsilon = -x'_b$, $\sum_{(i,j) \in T} x'_i x'_j \ge \sum_{(i,j) \in T} x_i x_j$, which contradicts the maximality of the number of $i$'s satisfying $x_i = 0$.
Therefore if $C := \{i : x_i > 0\}$, then $i, j \in C$ and $i < j$ implies $(i, j) \in T$. Thus if $C = \{i_1, i_2, \dots, i_k\}$ and $i_1 < i_2 < \dots < i_k$, then due to $i_j \ge 2i_{j-1}$ we have

$n > i_t \ge 2^{t-1}i_1 \ge 2^{t-1}$. This means that $t-1 \le \log_2 n$, or equivalently, $|C| \le k+1$. Applying the Cauchy-Schwarz inequality yields
$$\sum_{(i,j) \in T} x_i x_j = \frac{1}{2} \left( \left( \sum_{i \in C} x_i \right)^2 - \sum_{i \in C} x_i^2 \right) \le \frac{1}{2} \left( 1 - \frac{1}{|C|} \left( \sum_{i \in C} x_i \right)^2 \right) \le \frac{1}{2} \left( 1 - \frac{1}{k+1} \right).$$
In particular we have
$$M(n) \le \frac{\lfloor \log_2 n \rfloor}{2(\lfloor \log_2 n \rfloor + 1)},$$
which completes the proof. $\square$