If $x$ is a solution to the given equation, then $-x$ is also a solution, since
$|(-x)+|-x|+k|+|(-x)-|-x|-k|=2$
$|-x+|x|+k|+|-[x+|x|+k]|=2$
$|-[x-|x|-k]|+|x+|x|+k|=2$
$|x-|x|-k|+|x+|x|+k|=2$ which is the same as the original equation $|x+|x|+k|+|x-|x|-k|=2$.
Thus, in order for the equation to have an odd number of solutions, it is clear that $x=0$ must be one of these.
Substituting $x=0$ into the equation, we see that
$|k|+|-k|=2$
$|k|+|k|=2$
$2|k|=2$
$k|=\pm 1$
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[TR]
[TD]If $k=1$ and $x\ge 0$, then the equation becomes
$|2x+1|+1=2$ and so $x=0$
But $-x$ is a solution iff $x$ is, so there is only one solution in this case.[/TD]
[TD]If $k=-1$ and $x\ge 0$, then the equation becomes
$|2x-1|+1=2$ and so $x=0,\,1$
Thus, in this case, we have three solutions where $x=-1,\,0,\,1$
[/TD]
[/TR]
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$\therefore k=-1$ is the unique number for which the equation has exactly three solutions.