Let $X\in\mathbb{R}^{n}$ be a multivariate normal variable with the mean vector $\mu$ and the covariance matrix $\Sigma$. It is well known that if the matrix $\Sigma$ is positive-definite the following cumulative distribution \begin{equation} F\left(a_{1}, a_{2}, \cdots, a_{n}\right) = \mathbb{P}\left(X_{1}<a_{1},X_{2}<a_{2},\cdots,X_{n}<a_{n}\right) \end{equation} is monotonically increasing with respect to $a_{1},a_{2},\cdots,a_{n}$.
Is the above property still valid if the covariance matrix $\Sigma$ is not positive-definite (the degenerate case)?
The definition of a probability (and of a sigma algebra) necessarily implies that the CDF is always monotonically increasing with respect to any of its variables.
So, yes, it is also the case if Σ is degenerate.
Edit : A less theoretical answer consists of stating that if this was not the case, then you could find strictly negative probabilities. If $a_{1} < a_{1}^{'}$ and
\begin{equation} F\left(a_{1}, a_{2}, \cdots, a_{n}\right) > F\left(a_{1}^{'}, a_{2}, \cdots, a_{n}\right) \end{equation}
Then this means that
\begin{equation} \mathbb{P}\left(a_{1}\leq X_{1}<a_{1}^{'},X_{2}<a_{2},\cdots,X_{n}<a_{n}\right) < 0 \end{equation}
Edit : It seems I got tricked by the wording (I am not a native English speaker). The CDF is always nondecreasing, no matter what Σ is. If Σ is degenerate, then the PDF can be "not increasing". The simplest example is $ X=[X_{1}, X_{2}]$ where $X_1$ is a normal distribution and $X_2$ is $0$.