I'm pretty sure this doesn't follow Le Chatelier's Principle The likelihood is that I'm misunderstanding what's going on here.

Consider the reaction $\ce{A &lt;=&gt; B}$, where $K_\mathrm{c}=1$.


Initially, the system at equilibrium, where $[\ce A]=\pu{1M}$ and $[\ce B]=\pu{1M}$.
The system is then disturbed to increase $[\ce A]$ by $\pu{1M}$. Instantaneously, $[\ce A]=\pu{2M}$ and $[\ce{B}]=\pu{1M}$. Le Chatelier's principle suggests that the equilibrium position moves to the right to minimise the increase in $[\ce A]$.
The system then returns to equilibrium, such that $[\ce A]=\pu{1.5M}$ and $[\ce{B}]=\pu{1.5M}$.


Clearly, the relative amounts of $[\ce A]$ and $[\ce B]$ have not changed, so the equilibrium position has not changed, which is in direct contradiction with Le Chatelier's principle.

I think my misunderstanding may lie in one of the following:


My definition of equilibrium position (the relative amounts of reactants and products at equilibrium).
Between which numbers the equilibrium position is considered to move, let me elaborate on this: the relative amount of $[\ce A]$ and $[\ce B]$ have not changed between 1 and 3 ($1/1=1.5/1.5$), but they have changed between 2 and 3 ($1/2\ne 1.5/1.5$), so I see how the equilibrium position could be said to have moved right from step 2 to step 3 as the amount of B increases and the amount of A decreases.
Le Chatelier's Principle does not directly say what happens to concentration ratios.  Nor does it directly compare conditions after a disturbance with those before.  It says what happens only after a disturbance has been applied.

Here, the disturbance is adding one mole of A per liter.  When equilibrium is re-established you have only an additional one-half mole of A per liter, the rest having reacted to form B.  The initial addition of A has been partially offset, which is exactly what Le Chatelier's Principle says.