It's $\red{absolutely}$

We can perform the multiplications in the expression to get $|x^2+4x-21|>|x^2-4x-21|$

A trick we can use here is just squaring both sides; we get $\begin{aligned} \left(x^2+4x-21)^2>(x^2-4x-21 \right)^2 \\ x^4+8x^3-26x^2-168x+441 > x^4-8x^3-26x^2+168x+441 \\ 16x^3-336x > 0 \\ x(x^2-21) > 0 \end{aligned}$

Since we're looking for negative integer $x$ , we need to find negative integer solutions to $x^2-21<0$ (note the change in direction of the inequality, as we've divided by $x$ , which is negative). These are clearly just the integers $-4,-3,-2,-1$ with sum $\boxed{-10}$ .

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A couple of notes on this approach. It might seem like a crazy idea to choose to go from simple quadratics up to a quartic by squaring. First, the similarity in the two terms suggests there will be a lot of cancellation. Second, the usual method (which works fine) is to deal with the absolute value functions by considering different intervals; we can see the answer should be fairly simple; this again suggests that things should simplify fairly neatly, and we see that they do.

For $-3 \le x < 0$ , the inequality becomes:

$\begin{aligned} (-x+3)(x+7) - (x+3)(-x+7) & > 0 \\ - x^2 - 4x + 21 + x^2 - 4x - 21 & > 0 \\ - 8 x & > 0 \\ \implies - 3 \le x & < 0 \end{aligned}$

For $-7 \le x < -3$ , the inequality becomes:

$\begin{aligned} (-x+3)(x+7) - (-x-3)(-x+7) & > 0 \\ -x^2 - 4x + 21 - x^2 + 4x + 21 & > 0 \\ x^2 & < 21 \\ \implies - \sqrt{21} \le x & < - 3 \end{aligned}$

For $x < -7$ , the inequality becomes:

$\begin{aligned} (-x+3)(-x-7) - (-x-3)(-x+7) & > 0 \\ x^2 + 4x - 21 - x^2 + 4x + 21 & > 0 \\ 8x & \not > 0 & \small \red{\text{There is no solution.}} \end{aligned}$

Therefore the range of negative $x$ satisfying the inequality is $-\sqrt{21} \le x < 0$ and the sum of negative integer $x$ satisfying the inequality is $-4-3-2-1 = \boxed{-10}$ .

How I did it was:

$(x + 7) = -7$

$(x + 3) = -3$

$-7 + -3 = -7 - 3 = \fbox{-10}$