Millikan experiment: calculating $e$ Let's say we observed (or calculated) a series of measurements leading to a series of values for the charge of an oil-droplet which are grouped according to being $n\cdot e$, with $n\in\mathbb{N}$. Does it make a difference if you compute $e$ by taking the difference between two neighbouring groups or calculating the mean of each group and dividing it by the according $n$ respectively?
The only difference in my opinion is, that by taking the difference you cancel out systematic errors (e.g. values being both too big).
 A: As a matter of principle, it's nearly always easier to make a differential measurement directly than it is to measure two values absolutely and compute some small difference.  This is a theme that runs across many experiments, in many disciplines.
I am a strong proponent of reading classic papers directly: in general they are written quite clearly, in order to convince a skeptical audience of some then-new phenomenon.  In Millikan's paper, he confirms on the very first page that his technique is a differential measurement:

The essential feature of the method consisted in repeatedly changing the charge on a given drop by the capture of ions from the air and in thus obtaining a series of charges with each drop. These charges showed a very exact multiple relationship under all circumstances — a fact which demonstrated very directly the atomic structure of the electric charge.

The whole paper is shot through with techniques to remove systematic error by measuring ratios of things where the systematic issue cancels out. For instance, the paper begins by observing that the classical relationship between a droplet's radius $a$ and its terminal velocity when falling has to be modified if the radius is smaller than the mean free path $\ell$ for air molecules (because air molecules which miss the droplet don't contribute to any drag force).  The proposal for a first-order correction is
$$
v_\text{terminal} = \left(
\begin{array}{c}
\text{stuff about}
\\ \text{density and viscosity}
\\ \text{of air}
\end{array}
\right) \times \left(
1 + A \frac\ell a
\right)
\tag2
$$
where the dimensionless coefficient $A$ has to be determined from experiment.  However, for the landmark result, Millikan instead considers only ratios of terminal velocities for a single drop,
$$
\frac{v_1}{v_2} = \frac{mg}{qE-mg}
\tag6
$$
so that this $A$-bearing correction term cancels out.  (Millikan uses $F$ for electric field strength, but in contemporary notation we usually prefer $F$ for force and $E$ for field magnitude.)  He then solves this equation for the charge $q$ and re-writes his ratios of velocities as ratios of fall times. (By "fall time" Millikan means the time for the droplet to move a fixed vertical distance in the microscope. It doesn't matter what that distance is: it cancels out in the ratio.)  For a droplet which is observed to have a fall time under gravity alone $t_\text{grav}$, a fall time $t_E$ when the field is on and the droplet has charge $q=ne$, and a different fall time $t_E'$ for the same field when the droplet has charge $q'=(n+n')e$, Millikan constructs the relation
$$
n'e = \frac{mg}{E} t_\text{grav} \left( \frac1{t_E'} - \frac1{t_E} \right)
\tag{13}
$$
The purpose of this construction is to eliminate any first-order correction to the aerodynamics in the presence of the electric field: here we are subtracting two field-on measurements, rather than comparing a field-on to a field-off measurement.
In the section reporting his observations, Millikan writes

Since $n'$ is always a small number, and in some of the changes almost always has the value 1 or 2, its determination for any change is obviously never a matter of the slightest uncertainty. On the other hand, $n$ is often a large number, but with the aid of the known values of $n'$ is can always be found with absolute certainty so long as it does not exceed say 100 or 150.

From this follow data tables where Millikan describes the behavior of single droplets for as much as an hour at a time, with multiple charge states. Droplet number 6 enters the apparatus with charge $18e$, gains six charges to reach $24e$, loses seven charges to achieve $17e$, gains one to return to $18e$, and so on for a total of twelve different charge measurements.  Each of these is associated with a field-off fall-time measurement to demonstrate that the droplet's mass isn't changing.  (I learned reading this paper today that Millikan's apparatus included an x-ray tube so that he could ionize the air on command; my incorrect recollection was that he had to wait for the charge on a droplet to change stochastically.)
Each of the charge-state transitions in Millikan's data tables is associated with a computation of
$$
\frac{1}{n'} \left(
\frac1{t'_E} - \frac1{t_E}
\right)
= 
\frac1n \left(
\frac1{t_\text{grav}} + \frac1{t_E}
\right)
= 
\frac{eE}{mg} \frac{1}{t_\text{grav}}
$$
which is a constant for a given droplet, but which varies among different droplets.  Each droplet has a different mass $m$; also, this was before voltage regulators, so the electric field strength $E$ varied by about a half-percent each hour as the battery discharged.
The community was skeptical about Millikan's result in early versions of the experiment when he only measured the total charge on a droplet and looked for common denominators, as you suggest in your question. It was the ability to manipulate the charge on a single droplet, cancelling out systematic errors, which made the result convincing.
A: When you do Millikans experiment you don't know n or what two neighbouring groups are. The experiment shows, with enough droplets measured, that you get only multiples of some Number approximately 1.6*10^-19 As
to really measure e with some accuracy  this is not the best experiment. Once you have established, that its multiples of a number  you have a good estimate of the number with both methods.
