Show that for $X_1,X_2,\ldots,X_n$ i.i.d. standard normal,
$$\operatorname{Cov}\left(\bar{X}_n,\frac{1}{n}\sum|X_i-\bar{X}_n|\right)=0$$ where $$\bar{X}_n = \frac{1}{n}\sum X_i$$
What I do first is to write
$$\operatorname{Cov}\left(\bar{X}_n,\frac{1}{n}|X_i-\bar{X}_n|\right) = \frac{1}{n}E\left(\bar{X}_n \cdot\sum|X_i-\bar{X}_n|\right)-\frac{1}{n}E(\bar{X}_n)E\left(\sum|X_i-\bar{X}_n|\right)$$
Now, I now that the second part is zero, since $E(X_i)=0$ but then, I have no idea what to do with the absolute values in the first part. Any suggestions?
Since the expectation $\mathbb E\left(\bar{X}_n \cdot|X_i-\bar{X}_n|\right)$ exists, to prove that it equals zero it is sufficient to prove that $\bar{X}_n \cdot|X_i-\bar{X}_n|$ has symmetric distribution. Take $Y_i=-X_i$ and make sure that $Y_1,\ldots,Y_n$ are also i.i.d. standard normal, so the distribution of $\bar{Y}_n \cdot|Y_i-\bar{Y}_n|$ is the same as the distribution of $\bar{X}_n \cdot|X_i-\bar{X}_n|$, and $$ \bar{Y}_n \cdot|Y_i-\bar{Y}_n|=-\left(\bar{X}_n \cdot|X_i-\bar{X}_n|\right). $$