Dynamic range refers to the difference between the maximum signal a system can handle and the noise floor for a specific system configuration. If the gain of the system is fixed, there is no difficulty with this definition. If the gain of the system can be varied, then the dynamic range becomes a function of the system gain. The use of the word "dynamic" means that it refers to the system capability in processing a signal at one point in time, in other words for one specific gain. Since the gain can only be varied statically, you cannot claim increased dynamic range by having a variable gain. For example, if you have a 10-bit A/D converter in the system, the dynamic range is generally taken as approximately 6 dB X the number of bits or 60 dB in this case. If you put a variable gain amplifier (say with 0 to 40 dB gain in front of this A/D converter, you can't claim the dynamic range of the system is increased by 40 dB. The advantage of the variable gain amplifier is the ability to set the system dynamic range over a given voltage range. For example, if the maximum input of the A/D converter is 10 volts, then the dynamic range of the system with the variable gain amplifier can be shifted from 10 millivolts to 10 volts with the amplifier gain set to 0 dB, to 0.1 millivolts to 100 millivolts with the amplifier gain set to 40 dB. In both cases, the dynamic range remains at 60 dB but the static range has increased to 100 dB.
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.
Best Answer
A helpful way to think of the two concepts is suppose your final measured signal is a superposition of two functions:
$$ f = f_s + f_n $$ where \$f_s\$ is your signal, and \$f_n\$ is the noise.
The signal to noise ratio would be a measure the relative power of these two signals:
$$ SNR = \frac{P(f_s)}{P(f_n)} $$
Note that the SNR doesn't tell you anything about the "absolute" power of the noise; it is only a relative measure of how much stronger your desired signal is vs. the noise. You can have a lot of noise, but as long as your desired signal is much stronger you can still have a good SNR (think of someone using a megaphone in a really large crowd).
The noise floor is just a measure of the noise itself; that is, you can think of it as \$P(f_n)\$; how much noise do you have, regardless of whatever signal there is. This is like asking the question "how loud is the crowd I'm in?", and doesn't depend on any signal at all.