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In the comments below a post of mine, Glen_b and I were discussing how discrete distributions necessarily have dependent mean and variance.

For a normal distribution it makes sense. If I tell you $barx$, you haven't a clue what $s^2$ is, and if I tell you $s^2$, you haven't a clue what $barx$ is. (Edited to address the sample statistics, not the population parameters.)

But then for a discrete uniform distribution, doesn't the same logic apply? If I estimate the center of the endpoints, I don't know the scale, and if I estimate the scale, I don't know the center.

What's going wrong with my thinking?

EDIT

I did jbowman's simulation. Then I hit it with the probability integral transform (I think) to examine the relationship without any influence from the marginal distributions (isolation of the copula).

Data.mean <- Data.var <- rep(NA,20000)
for (i in 1:20000) 
 Data <- sample(seq(1,10,1),100,replace=T)
 Data.mean[i] <- mean(Data)
 Data.var[i] <- var(Data) 

par(mfrow=c(2,1))
plot(Data.mean,Data.var,main="Observations")
plot(ecdf(Data.mean)(Data.mean),ecdf(Data.var)(Data.var),main="'Copula'")

In the little image that appears in RStudio, the second plot looks like it has uniform coverage over the unit square, so independence. Upon zooming in, there are distinct vertical bands. I think this has to do with the discreteness and that I shouldn't read into it. I then tried it for a continuous uniform distribution on $(0,10)$.

Data.mean <- Data.var <- rep(NA,20000)
for (i in 1:20000)

 Data <- runif(100,0,10)
 Data.mean[i] <- mean(Data)
 Data.var[i] <- var(Data)


par(mfrow=c(2,1))
plot(Data.mean,Data.var)
plot(ecdf(Data.mean)(Data.mean),ecdf(Data.var)(Data.var))

This one really does look like it has points distributed uniformly across the unit square, so I remain skeptical that $barx$ and $s^2$ are independent.

It isn't that the mean and variance are dependent in the case of discrete distributions, it's that the sample mean and variance are dependent given the parameters of the distribution. The mean and variance themselves are fixed functions of the parameters of the distribution, and concepts such as "independence" don't apply to them. Consequently, you are asking the wrong hypothetical questions of yourself.

In the case of the discrete uniform distribution, plotting the results of 20,000 $(barx, s^2)$ pairs calculated from samples of 100 uniform $(1, 2, dots, 10)$ variates results in:

which shows pretty clearly that they aren't independent; the higher values of $s^2$ are located disproportionately towards the center of the range of $barx$. (They are, however, uncorrelated; a simple symmetry argument should convince us of that.)

Of course, an example cannot prove Glen's conjecture in the post you linked to that no discrete distribution exists with independent sample means and variances!

jbowman's Answer (+1) tells much of the story. Here is a little more.

(a) For data from a continuous uniform distribution, the sample mean
and SD are uncorrelated, but not independent. The 'outlines' of the plot emphasize the dependence.
Among continuous distributions, independence holds only for
normal.

set.seed(1234)
m = 10^5; n = 5
x = runif(m*n); DAT = matrix(x, nrow=m)
a = rowMeans(DAT)
s = apply(DAT, 1, sd)
plot(a,s, pch=".")

(b) Discrete uniform. Discreteness makes it possible to find a value $a$ of the mean and
a value $s$ of the SD such that $P(bar X = a) > 0,, P(S = s) > 0,$
but $P(bar X = a, X = s) = 0.$

set.seed(2019)
m = 20000; n = 5; x = sample(1:5, m*n, rep=T)
DAT = matrix(x, nrow=m)
a = rowMeans(DAT)
s = apply(DAT, 1, sd)
plot(a,s, pch=20)

(c) A rounded normal distribution is not normal. Discreteness causes
dependence.

set.seed(1776)
m = 10^5; n = 5
x = round(rnorm(m*n, 10, 1)); DAT = matrix(x, nrow=m)
a = rowMeans(DAT); s = apply(DAT, 1, sd)
plot(a,s, pch=20)

(d) Further to (a), using the distribution $mathsfBeta(.1,.1),$
instead of $mathsfBeta(1,1) equiv mathsfUnif(0,1).$
emphasizes the boundaries of the possible values of the sample mean
and SD. We are 'squashing' a 5-dimensional hypercube onto 2-space.
Images of some hyper-edges are clear.

set.seed(1066)
m = 10^5; n = 5
x = rbeta(m*n, .1, .1); DAT = matrix(x, nrow=m)
a = rowMeans(DAT); s = apply(DAT, 1, sd)
plot(a,s, pch=".")

Addendum per Comment.