Nice inequality ! We need two observations: 1) for every $a,b>0$ $a^3+b^3\ge a^2b+b^2a$ 2)if $a\le c$ and $b\le d$ then $a^2b+c^2d\ge a^2d+c^2b$ They follow easiy by rearangement. Now let's come back to our problem and use those $2$ observations to finish it: Let's rewrite our inequality as $$1012+\sum_{i \equiv j\pmod 2 } x_i^2x_j\ge \sum_{i \equiv 1+j\pmod 2} x_i^2x_j$$Now for every $i\equiv j\equiv 1 \pmod 2$ and $i<j$ we have that $x_i^2x_j+x_{i+1}^2x_{j+1}\ge x_i^2x_{j+1}+x_{i+1}^2x_j$.We sum this inequality for every $i\equiv j\equiv 1 \pmod 2$ and $i<j$ and so it is enough to prove that : $$1012\ge \sum_{i odd} x_i^2 x_{i+1}$$which is easy because $x_i^3+x_{i+1}^3\ge x_i^2 x_{i+1}+x_{i+1}^2 x_i\ge 2x_i^2 x_{i+1}$ and summing for $i$ odd and dividing by $2$ gives the result.

For $n > 1$, define \[f(x_1, x_2, \ldots, x_n) = \sum_{1 \leq i < j \leq n} (-1)^ix_i^2\cdot (-1)^jx_j.\]We want to prove that $f(x_1, \ldots, x_n) \geq -\sum x_i^3/2$ whenever $x_1 \leq \cdots \leq x_n$. First, note that, for $n - 1 > k > 1$. \[f(x_1, \ldots, x_n) = \sum_{1 \leq i < j \leq k} (-1)^ix_i^2\cdot (-1)^jx_j + \sum_{k + 1 \leq i < j \leq n}(-1)^ix_i^2\cdot (-1)^jx_j + \sum_{1 \leq i \leq k < j \leq n} (-1)^ix_i^2\cdot (-1)^jx_j,\]which can be simplified to \[f(x_1,\ldots, x_n) = f(x_1, \ldots, x_k) + f(x_{k + 1},\ldots, x_n) + \left(\sum_{1 \leq i \leq k}(-1)^ix_i^2\right)\left(\sum_{k + 1 \leq j \leq n}(-1)^jx_j\right).\]We are now ready to finish, using induction on $n$ (and the assumption $x_1 \leq \cdots \leq x_n$), where cases $n = 2, 3$ are easy: \[-x_1^2x_2 \geq \frac{x_1^3 + x_2^3}{3} \Longleftrightarrow 2x_1^2x_2 \leq x_1^3 + x_2^3,\]which is true, since $2x_1^2x_2 \leq x_1^2x_2 + x_1x_2^2$. For $n = 3$, note that \[x_1^2x_2 - x_1^2x_3 + x_2^2x_3 \leq x_2^2x_3 \leq \frac{x_2^3 + x_3^3}{2}.\]Now we do the inductive step. Since $f(x_1, \ldots, x_k) + f(x_{k + 1}, \ldots, x_n) \geq -\sum x_i^3/2$, it is enough to find $k$ such that \[\left(\sum_{1 \leq i \leq k}(-1)^ix_i^2\right)\left(\sum_{k + 1 \leq j \leq n}(-1)^jx_j\right) \geq 0.\]But the sign of $\sum_{k + 1 \leq j \leq n}(-1)^jx_j$ does not depend on $k$ since $x_1 \leq \cdots \leq x_n$ (it just depends on the parity of $n$), and $\sum_{1 \leq i \leq k}(-1)^ix_i^2$ only depends on the parity of $k$. Then, if $n$ is even, choose $k = 2$, while for $n$ odd choose $k = 3$ (this can be done since we assume $n \geq 4$ for $n$ even and $n \geq 5$ for $n$ odd here.)