Instrumentation Amplifier required for 24 bit ADC?
You say


  The signal will usually have a range between 10 and 70 uV. However, under certain circumstances the range can change, and I am expecting it to fluctuate between 10 and 500 uV (rare, but possible).


So, that's a range of 499 steps of 1µV between 1 µV and 500 µV. 

An ADC has \$2^N\$ steps, with \$N\$ being the number of bits.

You'd need 9 bits to represent 512 values, which would totally suffice. Your  expensive 24 bit ADC has an output of which you only care about the most significant 9 bit; 15 bit are totally irrelevant to you. Wrong choice to use a 24-bit ADC!

Most microcontrollers have a 10 bit or a 12 bit ADC built in. Go with that instead of using a dedicated ADC.

Then, you'd probably need to amplify the signal to match it to the range the built-in ADC. Since we don't know what that ADC is, we can't tell you what you'd need.

Also, as said before, you'd need to come clear about the signal source impedance. You're looking at a bandwidth of 40, maybe 50 kHz, so the source impedance will defined what you need in noise figure from your amplifier.

Simplifying a bit:

Thermal noise has a power of \$P = k_B B T\$; with \$B\$ being the bandwidth and \$T\$ being the temperature; \$k_B\$ is Boltzmann's constant.

At room temperature, this formula gives us \$-204\,\text{dBW}\$ (10⁻²⁰.⁴ watt per hertz). You're dealing with 50 kHz = \$\frac 12\cdot 10^5\text{ Hz}=47\text{ dBHz}\$, so you get \$(-204+47)\text{ dBW}=-157\text{ dBW}=10^{-15.7}\text{ W}\$ in noise power at the input of whatever amplifier you use.

The noise voltage is a function of the noise power and the input impedance of your amplifier:

\begin{align}
P &amp;= \frac{V^2}R\\
V &amp;= \sqrt{PR}\\
 &amp;= 10^{-7.85} \sqrt{\text W}\cdot \sqrt R
\end{align}

So, if your noise voltage needs to be lower than half a microvolt for your 1 µV requirement to even remotely make sense,

\begin{align}
V &amp;= 10^{-7.85} \sqrt{\text W}\cdot \sqrt R\\
&amp;\overset!&lt; 0.5\cdot 10^{-6}\text{ V}\\
\implies\\
\sqrt{R} &amp;&lt; \frac{0.5\cdot 10^{-6}\text{ V}}{10^{-7.85} \sqrt{\text W}}\\
&amp;= 0.5\cdot 10^{1.85}\frac{\text{V}}{\sqrt{\text W}}
\implies\\
R &amp;&lt; 0.25\cdot 10^{3.7}\text{ Ω}\\
&amp;\approx 1.25\text{ kΩ}
\end{align}

That means that even with the perfect, noise-free amplifier, your voltage source must be able to drive a 1.25 kΩ load, or your resolution is mathematically impossible to achieve (that's the worst kind of impossible).

Note that real-world amplifiers increase the noise. We measure that as Noise Figure, the ratio of signal-to-noise power ratio (SNR) coming out divided by SNR going in. Let's assume you'll have a non-trivial time building anything better than NF=3 dB.

As you can infer from above equations, this means that for your measurements to still make sense, you need to drop the input impedance by another 3 dB, i.e. half it, and your signal source still needs to drive that reliably.

So, that defines your amplifier needs – it's probably not going to be an instrumentation amplifier, as that solves few of the problems you have and gives you, as a cascade of multiple lower-gain amplifier stages, an additional noise figure problem.