A Sum for 2015

Algebra Level pending

Let \(x_1, x_2, x_3, \dots, x_{2015}\) be real numbers ranging from \(-1\) to \(1\) such that \(\sum_{i=1}^{2015}x_i^2=1\).

Find the remainder when the absolute value of the minimum possible value of \(\sum_{j=1}^{2015}x_j\sqrt{j^3}\) is divided by \(1000\).

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