Think Positive

First you should know that,
$|x| = \begin{cases} x \quad ; \quad x \geq 0 \\ -x \quad ; \quad x < 0 \end{cases}$

case : 1
When $x > 0$ $\dfrac{|x - |x||}{x} = \dfrac{|x - x|}{x} = 0$

case : 2
When $x = 0$ $\dfrac{0 - 0 }{0} = \text{indeterminate}$

case : 3
When $x < 0$ $\dfrac{ | x - (-x)|}{x} = \dfrac{|2x|}{x} = \dfrac{-2x}{x} = -2$

Therefore, $\dfrac{|x - |x||}{x} \ngtr 0 \ \forall \ x \in \Bbb R$

Hack: First, $0$ is not a solution since it makes the denominator $0$ . We know that the numerator will be a multiple of $x$ (modulus and subtraction don't change this fact, although it changes which multiple of $x$ is considered), which means if $x$ satisfies the condition, then so do $2x, 3x, 4x, \ldots$ . Thus the answer cannot be any finite positive integer; if it is, then there exists a solution $x$ , but we then know we have infinitely many solutions $x, 2x, 3x, 4x, \ldots$ . The answer cannot be negative either, since a count of something cannot be negative. Since the answer is presented as a "fill in the number" format, this answer must be finite; the two cases above rule out all but $\boxed{0}$ , the answer.

An actual solution has been provided elsewhere in this discussion.

\[ \displaystyle \frac { | x - | x | | } { x } = \begin { cases } x \ge 0 \\ x < 0 \end { cases } \rightarrow \text { if } x \ge 0, \text { then }, | x | = x \rightarrow \frac { | x - | x | | } { x } = 0 \text { otherwise, if } x < 0, \text { then }, | x | = -x \rightarrow \frac { | x - | x | | } { x } = \frac { | x + x | } { x } = \frac { | 2x | } { x } = \frac { -2x } { x } = -2 \text { because } x < 0, \text { not } x \le 0. \].