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.
 A: 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.
A: If the correlations between the three variables are $a$, $b$ and $c$, the eigenvalues of the correlation matrix satisfies
$$
\left|\begin{matrix}
1-\lambda & a & c \\
a & 1-\lambda & b \\
c & b & 1-\lambda
\end{matrix}\right|=0
$$
which after some algebra simplifies to
$$
(1-\lambda)^3 - (1-\lambda)(a^2+b^2+c^2)+2abc=0.
$$
For given values of $a$ and $b$, the correlation matrix is positive semi-definite when $c$ lies in a closed interval.  At the endpoints of this interval, one of the eigenvalues are zero implying that
$$
1-a^2-b^2-c^2+2abc=0.
$$
Solving this quadratic equation for $c$, we find that the endpoints of the interval of possible values of $c$ (plotted below) for given values of $a$ and $b$ are
$$
ab\pm \sqrt{(a^2-1)(b^2-1)}.
$$
If $a$ and $b$ are of opposite sign, the whole interval for $c$ thus contains only negative values if
$$
ab+\sqrt{(a^2-1)(b^2-1)}<0.
$$
Moving $ab$ to the right hands side and squaring both sides yields
$$
(a^2-1)(b^2-1)<a^2b^2
$$
which simplifies to the condition
$$
1<a^2+b^2.
\newcommand{corr}{\operatorname{corr}}
$$
Thus, to summarise, if $a=\corr(X,Y)$ and $b=\corr(X,Z)$ are of opposite signs and $(a,b)$ lies outside the unit circle, then $c=\corr(Y,Z)$ is necessarily negative.

