Reading the question and the comments, there may be a conceptual misunderstanding : the attenuator WILL attenuate any noise presented on its input (even from just a 50 ohm source impedance), to the same extent it attenuates the signal.
However it also generates noise of its own, which may be represented as the noise from a perfect resistor equal to its own output impedance, and this is added at the output to the (attenuated) input signal and noise. So if input and output Z are both 50 ohms, the net result is attenuated signal + marginally increased noise (i.e. NF = attenuation).
But if its output impedance is lower, the added noise is also lower, thus improving the noise voltage as Andy states.
So represent the attenuator as a perfect attenuator (attenuating noise) in series with a Johnson noise voltage source equal to the output impedance. The rest is just applying the formulae.
EDIT: re: updated question.
(1) There is nothing special about 290K except that it's a realistic temperature for the operation of a passive circuit. The reason they chose it is that the article quotes a noise floor ( -174dBm/Hz) which is correct for a specific temperature : yes, 290k.
(2) While any resistance in the attenuator will contribute noise, I realise that it is not a satisfactory explanation as to why you get the same noise out of an attenuator, because (as Andy says) you could make a capacitive attenuator which is not a Johnson noise generator. So we have to look a little deeper, and remember these noise sources are the statistics of the individual electrons that make up the current.
So, let's say we build a (50 ohm in, 50 ohm out) attenuator, and attempt to cheat Johnson by using a capacitive divider. That implies a node within the attenuator which conducts some of the input current to ground. At this node, we have two current paths; a fraction of the current flows to output, the rest to ground. What determines which path an individual electron will take? Essentially, chance. Collectively? Statistics. So this is a noise source.
Or let's just add series capacitance to provide enough attenuation : we thereby avoid dividing the current flow and eliminate the noise source, right? At the cost of reducing the signal current; our statistics now operate with a smaller sample size and consequently greater variance : more noise.
These results are the best you can do, there is no way round them.
Reading between the lines of the question: -
Firstly a signal that posseses energy has the potential to do work - it can convert that energy into heat or just pass that energy on to something else. Energy isn't lost, it either remains "potential" energy or gets transferred to something else partially or wholly.
When energy is taken from an object or a component (such as an inductor or capacitor or spring or flywheel), the amount of energy taken divided by the time elapsed to take that energy is power. Over a long time period this is often called "average power"
Therefore, for a signal possessing energy to have zero power I guess you could say that it is not losing or gaining any energy i.e. it is in equilibrium.
By contrast, if something is giving off power and is still in energy equilibrium it must have infinite energy.
I've done the best I can to try and read between the lines of this question and if you want a better answer you'll have to try and explain the statements in your question.
Best Answer
Your basic mistake is that power is energy per sample. Power is energy per time. In other words, P = E/t, not P = E/N as you used.
Resampling at a different rate doesn't change the time duration of the signal (t in the equation above). Resampling at a lower sample rate, for example, decreases the number of sample, but also increases the energy per sample.