Is the sum of two singular covariance matrices also singular? I have two sample covariance matrices, computed from $n$ samples, less than $p$ variables: they are singular then.
I know that the sum of two covariance matrices is also a covariance matrix.
My question is, if both are singular, is this sum also singular?
And if both are not singular, is this sum also not singular?

Sum of singular covariance matrices
No, a sum of singular matrices does not need to be singular.
See Is the sum of two singular matrices also singular?
A counter example in the answers sums two matrices that correspond to matrices for fully positive and fully negative correlation which are each singular but their sum is not.
$$\begin{bmatrix} 1&1\\
1&1
\end{bmatrix}
+
\begin{bmatrix} 1&-1\\
-1&1
\end{bmatrix} = \begin{bmatrix} 2&0\\
0&2
\end{bmatrix} $$
Sum of non-singular covariance matrices
In general it is possible for two non-singular matrices to add up to a singular matrix. E.g. for any non-singular matrix $A$ the matrices $A$ and $B=-A$ are non-singular but the sum $A+B=0$ is singular.
However, as mentioned in the comments this is not true for covariance matrices which are positive definite of the variables are independent. The sum of positive definite matrices, which are non-singular, are positive definite and remain non-singular.
Intuitive approach
If some matrix is a covariance matrix then it has a square root and can be written as $X^tX$. From the definition of the covariance matrix, it is the cross product of vectors after their mean is subtracted.
Then the sum of two covariance matrix can be seen as a single matric where the vectors are concatenated.
The property of singularity can be linked to the independence of the vectors in $X$. As you say, if $X$ has less samples $n$ than variables $p$ then the matrix will be singular. But by adding the samples of two variables together you can get that the variables become independent and the matrix will be non-singular. The other way around can not happen. Once the variables are independent, then you can not get then dependent by adding more samples.
With the example above
$$\overbrace{
\begin{bmatrix} 1\\1
\end{bmatrix}\cdot
\begin{bmatrix} 1&1\\
\end{bmatrix}
+
\begin{bmatrix} 1\\-1
\end{bmatrix}\cdot
\begin{bmatrix} 1&-1
\end{bmatrix}}^{\begin{bmatrix} 1&1\\
1&1
\end{bmatrix}
+
\begin{bmatrix} 1&-1\\
-1&1
\end{bmatrix} }
= \overbrace{
\begin{bmatrix} 1&-1\\
1&1
\end{bmatrix}\cdot
\begin{bmatrix} 1&1\\
-1&1
\end{bmatrix}}^{
  \begin{bmatrix} 2&0\\
0&2
\end{bmatrix} }$$
