Solution:

Since $\mathcal{T}$ shares a side with each of the other rhombuses, all five have the same side length $z.$ Let $Y$ be the vertex of $\mathcal{T}$ on $\overline{AB},$ and let $\alpha$ be the angle of $\mathcal{P}$ at $Y.$ Then each congruent rhombus has area $z^2 \sin\alpha = \sqrt{2006}.$ The angles of $\mathcal{P},$ $\mathcal{T},$ and $\mathcal{Q}$ at $Y$ lie along the line $AB,$ and by symmetry $\mathcal{Q}$'s angle there also equals $\alpha,$ so $\mathcal{T}$'s angle is $180^\circ - 2\alpha.$ Hence $$K = z^2 \sin(180^\circ - 2\alpha) = z^2 \sin 2\alpha = 2 z^2 \sin\alpha \cos\alpha = 2\sqrt{2006}\,\cos\alpha.$$

As $\alpha$ ranges over $(0^\circ, 90^\circ),$ the value $\cos\alpha$ takes every value in $(0, 1),$ so $K$ takes every value in $\left(0, \sqrt{8024}\right).$ Since $89^2 = 7921 \lt 8024 \lt 8100 = 90^2,$ the possible positive integer values are $1, 2, \ldots, 89:$ there are $89$ of them.