How do you prove Cov $\left( \bar{X} , X_i - \bar{X} \right) = 0$? How do you prove Cov $\left( \bar{X} , X_i - \bar{X} \right) = 0$ given $ X_1 ,..., X_i$ are i.i.d. each with variance $\sigma^2$ and $\bar{X}$ is the sample mean?? In other words, how do you show that the sample mean and the differences of the observations from that mean are not linearly correlated???
Am I on the write track? 
I've written
$Var \left( \bar{X} + \left( X_i - \bar{X} \right) \right) = Var  X_i =  Var \bar{X} + Var \left( X_i - \bar{X} \right) + Cov \left( \bar{X} , X_i - \bar{X} \right)$ . 
I know 
$Var  X_i = \sigma^2 , Var \bar{X}= \frac{\sigma^2}{n}$, so all I need left to find the covariance is $Var \left( X_i - \bar{X} \right)$. 
I don't know if this is right, but I wrote
$Var \left( X_i - \bar{X} \right) = E\left( \left( \left( X_i - \bar{X} \right) - \mu_{ X_i - \bar{X}} \right)^2 \right)=E \left( \left( X_i - \bar{X} \right)^2 \right)=s^2$, but I don't know if that's right or helpful at all. 
Am I on the right track??
 A: Hint: Use properties of the covariance function and independence of $X_1,...,X_n$.
Solution:
Note that
$$ Cov(\bar{X},X_i-\bar{X}) = Cov(\bar{X},X_i)-Cov(\bar{X},\bar{X}) = Cov(\bar{X},X_i)-Var(\bar{X}) $$
You know that $Var(\bar{X})=\frac{\sigma^2}{n}$. So you need to find $Cov(\bar{X},X_i)$. But,
$$ Cov(\bar{X},X_i)= Cov(\frac{1}{n}\sum_{j=1}^{n}X_j,X_i) = \frac{1}{n} \sum_{j=1}^{n}  Cov(X_j,X_i) $$
Since $X_j$'s are independent it follows that
$$Cov(X_j,X_i)=0 \text{ for all j } \neq i $$
So we conclude that
$$ Cov(\bar{X},X_i) = \frac{1}{n} Cov(X_i,X_i) = \frac{\sigma^2}{n} $$
It follows that 
$$Cov(\bar{X},X_i-\bar{X}) = 0$$
A: 
Note: This answer was originally posted in response to a duplicate Question, now closed. The first part is similar to the excellent answer of @Mdoc (+1). The second part discusses the importance of the result in statistical inference.

Residuals about a mean have $0$ covariance with the mean.
Without loss of generality, find $Cov(X_1-\bar X, \bar X):$
Then
$$Cov(X_1 - \bar X, \bar X) = Cov(X_1, \bar X) - Cov(\bar X,\bar X)\\ = Cov(X_1, \bar X) + Var(\bar X) = Cov(X_1,\bar X) -\sigma^2/n.$$
Now
$$Cov(X_1,\bar X) = Cov\left(X_1, \frac 1n\sum_{i=1}^nX_i\right)\\
=Cov\left(X_1,\frac 1n X_1\right) + 0 = \frac 1n Cov(X_1,X_1)\\
= \frac 1n Var(X_1) = \sigma^2/n.$$
Thus, $Cov(X_1,\bar X) = \sigma^2/n - \sigma^2/n = 0.$
Relavance to statistical inference. This result is important in statistical inference. The residuals $r_i = X_i - \bar X$
of observations from their group means are widely used in ANOVA and regression.
Sample mean and variance independent for normal data. For normal data uncorrelated implies independent. Because $\bar X$ is independent of the $r_i,$ then it is independent of $S.$ So for normal data $\bar X$ and $S_X^2$ are stochastically independent. (They are not 'functionally' independent because $\bar X$ is used to find $S_X^2.)$ This is important for t statistics because Student's t distribution is
defined in terms of a ratio with numerator and denominator independent.
Simulations illustrating lack of correlation. A brief simulation in R illustrates that means are not correlated with residuals from them.
(The simulation uses 10 million normal samples
of size $n=10,$ giving several decimal places
of accuracy for the correlation.)
set.seed(2020)
M = 10^7; n = 10
X = rnorm(M*n, 100, 15)
DTA = matrix(X, nrow=M)
A = rowMeans(DTA)
X1 = DTA[,1]
cor(X1-A,A)
[1] -0.0004722208  # aprx 0

A similar simulation with exponential data also shows lack of correlation:
set.seed(2020)
M = 10^7; n = 10
Y = rexp(M*n)
DTA = matrix(Y, nrow=M)
A = rowMeans(DTA)
Y1 = DTA[,1]
cor(Y1-A,A)
[1] 4.620507e-08

However, scatterplots of residuals against means illustrates independence for the normal data, but a clear pattern of dependence for the exponential data. (We use reduced numbers of datasets for a manageable number of points in the scatterplots.)

m=30000
x1=X1[1:m]; a.x=A[1:m]; r.x=x1-a.x
y1=Y1[1:m]; a.y=A[1:m]; r.y=y1-a.y
par(mfrow=c(1,2))
 plot(a.x,r.x, pch=".", main="Normal Data")
 plot(a.y,r.y, pch=".", main="Exponential Data")
par(mfrow=c(1,1))

