Let $X_1,X_2$ follow a bivariate standard normal distribution with some non-zero correlation coefficient, $\rho\neq 0$. Let the function $f(z) = z^k,\; k=1,...$. By Stein's lemma, we have that
$$\text{Cov}(f(X_1),X_2) = \text{Cov}(X_1,X_2)\cdot E[f'(X_1)]$$
and in our case
$$\text{Cov}(X_1^k,X_2) = E[X_1^k X_2] = \rho\cdot kE(X_1^{k-1})$$
Since $E(X_1^{k-1}) = 0$ when $k-1$ is odd, it follows that
$$\text{Cov}(X_1^k,X_2) = E[X_1^k X_2] = 0,\;\; \text{iff}\;\;\ k+1\;\;\text{is odd}$$
namely when the sum of the moment-orders is odd.
Moreover by Isserlis Theorem, if $X_1,...,X_n$ follow jointly a zero-mean multivariate normal distribution with dependence, we have for $m\leq n$
$$E[X_1\cdot ... \cdot X_m] = 0 \;\; \text{iff}\;\;\ m\;\;\text{is odd}$$
What I cannot seem to find anywhere is the generalized combination of the two results and its proof, namely that if $X_1,...,X_n$ follow jointly a zero-mean multivariate normal distribution with dependence, we have
$$E[X_1^{r_1}\cdot ... \cdot X^{r_m}_m] = 0 \;\; \text{iff}\;\;\sum_{j=1}^m r_j \;\;\text{is odd}$$
Do we?