Why does "V1 increase with decreasing density"?
I guess the statement is about V1 from the balanced field length, which is where the accelerate-go distance is equal to the accelerate-stop distance:

(image source: boldmethod &gt; Engine Failure On Takeoff: Do You Stop Or Go?)
How do these curves change when the density decreases? Lower density means less thrust, therefore accelerating will be slower and uses more distance. This affects both curves equally for the acceleration to V1, but afterwards only the accelerate-go distance:

Let us call the force accelerating the aircraft initially $ F = 2 \times T $ (thrust from two engines).
In the accelerate-go case, you would only accelerate with a force of $ F = T $ after the engine failure.
In the accelerate-stop case, you brake with a certain force $ F = F_\text{brake} $.
Now, at lower density, the thrust $ T $ will decrease, but the brake force $ F_\text{brake} $ will be unaffected.
The accelerate-go case is therefore affected during the entire time by the lower value of $ T $.
The accelerate-stop case is only affected during the accelerate part, but afterwards the brake force is still the same.

Therefore, both curves will move up in the diagram, but the accelerate-go distance will move up more. This implies that their intersection (the balanced field length) will occur further to the right, resulting in a higher value for V1.