Can there be more than one combination of reactant/product concentration for a specific equilibrium constant? For an elementary reaction like $$\ce{A + B -&gt; AB},$$ $$K_{c} = \frac{\ce{[AB]}}{\ce{[A][B]}}$$ and $$\frac{\mathrm{d}\ce{[AB]}}{\mathrm{d}t} = k\ce{[A][B]}.$$ If I let the reaction reach equilibrium, and at this point, without altering temperature, suck away $\ce{A}$ and add $\ce{B}$ at the same rate (so $-\mathrm{d}\ce{[A]}/\mathrm{d}t = \mathrm{d}\ce{[B]}/\mathrm{d}t$) and alter both concentrations slightly, once I am finished, will the system still be in equilibrium or try to shift? 

What if the reaction is not an elementary reaction such that $\frac{\mathrm{d}\ce{[AB]}}{\mathrm{d}t} = k\ce{[A]}^n\ce{[B]}^m$?
A single "elementary reaction" does not involve an equilibrium.  

There must be a forward reaction and a reverse reaction to have an equilibrium.  This is at least two elementary reactions.  

Considering an equilbrium reaction: Just as a square encloses the most area for a rectangle of given perimeter, if the sum [A] + [B] is kept constant,  [A][B] is maximal when [A] = [B].  Deviation from this maximum will shift the equilibrium away from the product and toward the reactants.  In other words, if the system is at equilibrium with [A] = [B], and [A] is decreased and [B] increased, keeping [A] + [B] constant, there will be net reverse reaction until equilibrium is again reached, and [AB] at equilibrium will be less than orginally.