Does $Cov(X,Y)<0$ and $Cov(X,Z)>0$ imply $Cov(Y,Z)<0$? Consider three normally distributed random variables, $X,Y,Z$ where $Cov(X,Y)<0$ and $Cov(X,Z)>0$. Can we say anything about the sign of $Cov(Y,Z)$?
Intuitively, $Y$ goes down when $X$ goes up. Next, $Z$ goes up when $X$ goes up. I would guess $Cov(Y,Z)<0$.

Set $\rho=Corr(X,Y)<0$. Then, $Y=\rho X+\sqrt{1-\rho^2}\epsilon$ where $\epsilon\sim N(0,1)$ is orthogonal to $X$. Then,
\begin{align}
Cov(Y,Z) &= Cov\left(\rho X+\sqrt{1-\rho^2}\epsilon,Z\right) \\
&=\underbrace{\rho Cov(X,Z)}_{<0} + \sqrt{1-\rho^2}Cov(\epsilon,Z).
\end{align}
Is there an argument why $Cov(\epsilon,Z)$ should be zero or negative? Or are additional assumptions necessary?

Let's write $Z=r X+\sqrt{1-r^2}\epsilon_Z$ where $\epsilon_Z\sim N(0,1)$ is orthogonal to $X$.. Then,
\begin{align}
Cov(\epsilon,Z) = \sqrt{1-r^2}Cov(\epsilon,\epsilon_Z).
\end{align}
I cannot see why $Cov(\epsilon,\epsilon_Z)$ should be zero/negative.

No (to the title question). Any positive semi-definite matrix is also a covariance matrix. So, if there exists a $\rho$ that the following matrix is PSD, then you've a contradiction:
$$\Sigma=\begin{bmatrix}1&-\rho&\rho\\-\rho &1&\rho\\\rho&\rho&1\end{bmatrix}$$
This occurs for $\rho=0.1$ for example. The conditions on when it implies are correctly laid out in Jarle’s answer.
