Let $A$ be an $n\times n$ symmetric matrix. Then, $A$ is a positive semidefinite iff every principal minor of $A$ is $\geq0$; $A$ is a negative semidefinite iff every principal minor of odd order is $\leq0$ and every principal minor of even order is $\geq0$.
Now, let $B$ be a $4\times 4$ symmetric matrix with
- principal minor of odd order $=0$ and principal minor of even order $\geq0$.
- all principal minors $=0.$
I think in both the cases we have positive as well as negative semidefinite matrices, but I think how a single matrix be both +ve and -ve?
For all principal minors $=0$, you got your answer in Yurii's comment: such matrix is equal to the zero matrix, which is both positive and negative semidefinite.
As for the case
consider all principal minors of order $1$. They are equal to zero, which means that the diagonal elements of $B$ are equal to zero.
Now consider all principal minors of order $2$: they are $\ge 0$. But, these are determinants of symmetric matrices of order $2$ with zero diagonal, so
$$0 \le \det \begin{bmatrix} 0 & b_{ij} \\ b_{ij} & 0 \end{bmatrix} = -b_{ij}^2,$$
so $b_{ij} = 0$ for all $i \ne j$. In other words, $B = 0$, as in your other case.
Even without this analysis, the answer is quite obvious: by your condition, $B$ is both positive and negative semidefinite. It is easy to show that this is possible if and only if $B = 0$.