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Note that $E=\frac{1}{2}\sum_{cyc}(x_i-x_{i+1})^2$. It is clear that for even $n$, $\text{sup}(E)=\frac{n}{2}$, which is achieved by any cyclic permutation of $(1,0,1,0,...,1,0)$. For odd $n$, note that $(x_1, x_2,..., x_n)=(1,0,...,1)$ gives $E(1,0,...,1)=\frac{n-1}{2}$. This is in fact the supremum, since each variable locally maximizes the squares by being either $0$ or $1$. The supremum is achieved by any cyclic permutation of $(1,0,1,0,...,1,0,0)$ or $(1,0,1,0,...,1,0,1)$ in this case.

It is an easier version of 2011 KJMO P7 See also 2011 KMO P8