We want that
$$\sum_{1 \le i < j \le 2024} (-1)^{i+j} x_i^2 x_j \ge -1012.$$
Now, observe that the LHS is
$$- \left( \sum_{i=1}^{1012} x_{2i-1}^2 x_{2i} \right) + \sum_{i=1}^{1012} \left( (x_{2i}^2 - x_{2i-1}^2) \left( \sum_{j<i \le 1012} (x_{2j} - x_{2j-1}) \right) \right).$$
But, $x_{2i}^2 \ge x_{2i-1}^2$ and $x_{2j} \ge x_{2j-1}$, thus we will be done if $\sum_{i=1}^{1012} x_{2i-1}^2 x_{2i} \le 1012$. But,
$$3x_{2i-1}^2 x_{2i} \le 2x_{2i-1}^3 + x_{2i}^3 \le 3 \cdot \frac{x_{2i-1}^3 + x_{2i}^3}{2}.$$
Thus, we just need $\sum_{i=1}^{1012} \frac{x_{2i-1}^3+x_{2i}^3}{2} \le 1012$ but this is precisely the given condition so we are done!

Solution:
We use induction on $k$ to show the following: given non-negative reals $x_1 \le x_2 \le \dots \le x_{2k}$, we have
$$\sum_{i=1}^{2k} x_i^3 + 2 \sum_{1 \le i < j \le 2k} (-1)^{i+j} x_i^2 x_j \ge 0$$
with equality iff $x_{2i-1} = x_{2i}$. This suffices since $x_1^3 + x_2^3 + \dots + x_{2024}^3 = 2024$.
Note that the left-hand side can be written as $x_{2k}^3 - 2cx_{2k} + d$, where
$$c = x_{2k-1}^2 - x_{2k-2}^2 + \dots - x_2^2 + x_1^2$$
and
$$d = \sum_{i=1}^{2k-1} x_i^3 + 2 \sum_{1 \le i < j \le 2k-1} (-1)^{i+j} x_i^2 x_j \ge 0$$
are independent of $x_{2k}$. Consider the function $f(x) = x^3 - 2cx + d$. Its derivative is $3x^2 - 2c$, and so $f$ is increasing as long as $x^2 \ge 2c/3$. However,
$$c = x_{2k-1}^2 + (-x_{2k-2}^2 + x_{2k-3}^2) + \dots + (-x_2^2 + x_1^2) \le x_{2k-1}^2,$$
so either $c < 0$, in which case $x_{2k-1}^2 \ge 0 > 2c/3$, or $c \ge 0$, in which case $x_{2k-1}^2 \ge c \ge 2c/3$.
Thus we must have
$$f(x_{2k}) \ge f(x_{2k-1}) = \sum_{i=1}^{2k-2} x_i^3 + 2 \sum_{1 \le i < j \le 2k-2} (-1)^{i+j} x_i^2 x_j$$
and the rest follows by induction. $\square$