Positive definiteness of sample covariance matrix when $ N < p $ Say I have a sample where $ N < p $, where $ N $ denotes the number of observations and $ p $ the number of variables. I know that the rank of the covariance matrix is then at most $ N $. I know that it is not positive definite since it is not invertible. However, I read from Wikipedia that a covariance matrix has to be at least positive semidefinite. Can I deduce positive semidefiniteness from here?

Yes, the sample covariance matrix will still be positive semi-definite.
To see this, note that if $X\in \mathbb{R}^{N\times p}$ is the data matrix (with observations in the rows and variables in the columns), then the sample covariance matrix is $C := \frac{1}{N-1}Y^TY\in \mathbb{R}^{p\times p}$, where $Y\in \mathbb{R}^{N\times p}$ is the matrix $X$ with each column's mean subtracted from that column's entries.
Note that $C$ is symmetric as $C^T = \left( \frac{1}{N-1}Y^TY\right)^T = \frac{1}{N-1}Y^TY=C$.
Also, for any $\mathbf{v}\in\mathbb{R}^{p}$, we have
$$\begin{align*}
\mathbf{v}^T C \mathbf{v}&= \frac{1}{N-1}\mathbf{v}^TY^TY\mathbf{v}\\
&=  \frac{1}{N-1} \left( Y\mathbf{v}\right)^T Y\mathbf{v}\\
&= \frac{1}{N-1}\left\| Y\mathbf{v}\right\|^{2}\\
&\ge 0.
\end{align*}
$$
Thus the sample covariance matrix $C$ is positive semi-definite.
