Le Chatelier Question

[turtlebanana]

For the reaction which is existing in equilibrium, in a closed vessel and at SLC:
2NO₂ (g) (brown)  ⇆  N₂O₄ (g) (colourless)

If additional NO2 was added, then according to Le Chatelier's principle, wouldn't this cause a net forward reaction, producing more N₂O₄ particles, resulting in an increase its concentration and thus causing the colour of the mixture to become more colourless?

The answer claims the colour of the mixture becomes more brown and i can see why it would be logical to think so - because more reactant is being added, thus increasing the concentration of NO₂.
But then doesn't Le Chatelier's principle come into play - meaning that the reaction would shift forward, thus causing the colour of the mixture to become more colourless instead of more brown?

[Sine]

For the reaction which is existing in equilibrium, in a closed vessel and at SLC:
2NO₂ (g) (brown)  ⇆  N₂O₄ (g) (colourless)

If additional NO2 was added, then according to Le Chatelier's principle, wouldn't this cause a net forward reaction, producing more N₂O₄ particles, resulting in an increase its concentration and thus causing the colour of the mixture to become more colourless?

The answer claims the colour of the mixture becomes more brown and i can see why it would be logical to think so - because more reactant is being added, thus increasing the concentration of NO₂.
But then doesn't Le Chatelier's principle come into play - meaning that the reaction would shift forward, thus causing the colour of the mixture to become more colourless instead of more brown?

Le Chatelier's suggests that the system attempts to bring the system back into equilbrium. In doing so it doesn't necessarily fully revert the effects. So if you are adding in the brown reagent the system will try to go with the forward reaction (to make it colourless) but it won't be enough to get it back to the original colour. A mathematical approach to this would probably make it a bit more clearer.