Number of lenslets in wavefront sensor array I have to do a short calculation, but, quite frankly, I have no idea how to even start...
Suppose you have a 10-m telescope (f/10) with an infrared camera that observes at 10 micrometer. The seeing is 2 arcsec (FWHM of the PSF,  2”= 1/100000 rad) and you would like to implement an adaptive optics system.
The question is: How many lenslets do you need in the Shack-Hartmann wavefront sensor array? Below is a picture of the working principle of such an array. The telescope would also include a coronagraph, but I think (might be wrong) you don't need that information for this calculation.
Generally speaking as many as possible because the number of lenslets determines the lateral wavefront resolution. But in reality there are a few factors to be considered.
Wavefront sensor
Say your lenslet array is of focal length $f$ and subaperture size $D$, given a maximum desired detection angle $\alpha$ (in your case is $2'' \approx 1\times10^{-5}\, \text{rad}$), there is the following relationship:
$$
2'' = \alpha \approx \tan \alpha = \frac{\Delta x}{f} \leq \frac{D}{2f}. \tag{1}
$$

To ensure a good resolution and reduce cross-talk effect, $D$ is chosen to be at least a few image sensor pixel size. In other word, for a given image sensor to measure the seeing angle, the number of lenslets are limited to a maximum value.

Usually you do not design the lenslet arrays; those are pre-designed, e.g. https://www.thorlabs.com/newgrouppage9.cfm?objectgroup_id=2861. So Eq. (1) serves as a metric to select between different microlens array designs.

For AO applications, the primary goal of wavefront sensors is the measurement speed rather than the lateral resolution. The sensors need to be fast. The smaller the number of pixels, the smaller the sensor latency is, resulting in a low number of pixels, and hence low number of lenslets.


Adaptive optics (AO) system
$w \in \mathbb{R}^m$: Measurement wavefront of a lenslet array of number $m$.
$c \in \mathbb{R}^n$: Correction wavefront of the deformable mirror that has $n$ number of actuators.
An adaptive optics (AO) system works by solving a linear system to output $c$ given $w$:
$$
w = A c
$$
where interaction matrix $A \in \mathbb{R}^{m \times n}$ is pre-calibrated before usage.

To compute $c$ in a least squares sense, it requires $m \geq n$, meaning an over-determined system, mentioned also by @nflemming2004.

As for the aperture sizes of the two, it does not really matter that much because the conjugate lenses/mirrors in your optical system can scale down/up the apertures. That said, the sub-aperture of the wavefront sensor (i.e. $D$) and the actuator cover area $D_{\text{deformable_mirror}}$ do not have to be equal.

If $m \gg n$ (as in most cases), as you can imagine, the extra number of lenslets only serves to make a robust estimate, and is wasted from AO's perspective because of insufficient correction degree-of-freedom.


Example
Using WFS30-7AR as our wavefront sensor, that $f = 5.6 \,\text{mm}$ and $D  = 150\,\text{um}$, given Eq. (1) we know:
$$
2'' \leq \frac{D}{2 f} = 1.34 \times 10^{-2} \, \text{rad} \tag{satisfied}
$$
The number of lenslet is $m = 80 \times 80$, far larger than $n = 140$ the actuator number of deformable mirrors (say https://www.thorlabs.com/newgrouppage9.cfm?objectgroup_id=3208).
So for this design, we conclude it satisfies the requirement of this question.