Johnson noise potential divider
It's the noise of a 20k\$\Omega\$ resistor.  

edit
The first paragraph is a bit ambiguous to me. The 20k\$\Omega\$ is when you measure across both resistors (that's what I thought you meant, since you mention the 20k\$\Omega\$). If, however, you would measure across one of the resistors you'd have to see both in parallel, and then it would become the noise of a 5k\$\Omega\$ resistor.

But noise can't be added arithmetically because of its stochastic character. Noise voltage is defined as  

\$ v = \sqrt{4kT R\Delta f} \$  

so it's proportional to the square root of the resistance. Therefore the noise of a resistor of 20k\$\Omega\$ is \$\sqrt{2}\$ higher than that of a 10k\$\Omega\$ resistor, not twice as high. Placing the resistors in parallel will cause the noise voltage to be \$\sqrt{2}\$ lower.  

I've seen a couple of designs where this property was used to improve the noise of an opamp by placing two of them in parallel. But this will only improve noise figures by -3dB. Barry also mentions this. Probably not worth it.
A few years ago Elektor published a power amplifier design based on 32 parallel NE5532s. Then it becomes interesting. The noise is reduced by 15 dB, and the specs for the amplifier show a S/N ratio of 110dBA.  

edit
For the two resistors in series with the other end of the second resistor open: that resistor's noise voltage doesn't count. The open end would show some Johnson noise with respect to the common node, but the other resistor won't see it. (The noise it would pick up as an antenna is not the resistor's Johnson noise.)