From the definition of the covariance matrix,
$${\sigma}_{ij}=\left\langle \frac{x_i x_j+ x_j x_i}{2}\right\rangle -\langle x_i\rangle\langle x_j\rangle,$$ where we define
$$\boldsymbol{x}=(q_1,p_1,\cdots, q_N, p_N)^\top.$$
Given that $2i\Omega_{jk}=[x_j,x_k],$ we learn that
$${\sigma}_{ij}=\left\langle x_i x_j\right\rangle -\langle x_i\rangle\langle x_j\rangle-i \Omega_{ij}\quad\Rightarrow \quad \sigma_{ij}+i\Omega_{ij}=\left\langle x_i x_j\right\rangle -\langle x_i\rangle\langle x_j\rangle=\left\langle\left(x_i-\left\langle x_i\right\rangle\right)\left(x_j-\left\langle x_j\right\rangle\right)\right\rangle.$$
Next, the right hand side must be greater than or equal to zero due to standard properties of covariance matrices. We have that
$$\boldsymbol{\sigma}+i\boldsymbol{\Omega}=\langle (\boldsymbol{x}-\langle\boldsymbol{x}\rangle)(\boldsymbol{x}-\langle\boldsymbol{x}\rangle)^\top\rangle\equiv\langle\boldsymbol{X}\boldsymbol{X}^\top\rangle.$$ We can follow a simple proof using an arbitrary real-valued vector $\boldsymbol{w}$:
$$\boldsymbol{w}^\top \langle\boldsymbol{X}\boldsymbol{X}^\top\rangle\boldsymbol{w}= \langle \boldsymbol{w}^\top\boldsymbol{X}\boldsymbol{X}^\top\boldsymbol{w}\rangle= \langle \left(\boldsymbol{w}^\top\boldsymbol{X}\right)^2\rangle= \langle \left|\boldsymbol{w}^\top\boldsymbol{X}\right|^2\rangle\geq 0.$$ In the final inequality we used that each of the $x_i$ is Hermitian.
Any matrix $\boldsymbol{M}$ that satisfies $\boldsymbol{w}^\top\boldsymbol{M}\boldsymbol{w}\geq 0$ for all real vectors $\boldsymbol{w}$ is by definition positive semidefinite. Thus
$$\boldsymbol{\sigma}+i\boldsymbol{\Omega}\geq \boldsymbol{0}\boldsymbol{0}^\top.$$
This relation essentially translates between the symmetrized version of the covariance $\boldsymbol{\sigma}$ to the standard non-symmetrized version of the covariance matrix $\langle\boldsymbol{X}\boldsymbol{X}^\top\rangle$.
To answer the question in the title, all states have their covariance matrices satisfy $\boldsymbol{\sigma}+i\boldsymbol{\Omega}\geq 0$. This is not only a property of Gaussian states! The property of Gaussian states is that they are completely characterized by their covariance matrices (and their mean value vectors $\langle \boldsymbol{x}\rangle$).
So maybe you want to ask why one only needs to specify the covariance matrix if one wants to completely describe a Gaussian state? The answer is given by Isserlis's theorem (sometimes more familiar in terms of Wick's theorem). To completely describe any state, one could list all of its moments $\langle x_i x_j\cdots\rangle$. Given knowledge of the mean values $\langle\boldsymbol{x}\rangle$, we can instead describe a state by all of its zero-mean moments $\langle X_i X_j\cdots\rangle$. The Isserlis theorem says that, for Gaussian probability distributions, all of the higher-order moments can be found from sums of products of all of the different pairings of the the second-order moments (ie, elements of the covariance matrix), such as \begin{equation}\langle X_i X_j X_k X_l\rangle = \langle X_i X_j\rangle \langle X_k X_l\rangle+\langle X_i X_k\rangle \langle X_l X_l\rangle + \langle X_i X_l\rangle \langle X_k X_j\rangle .\end{equation} Gaussian state $\Rightarrow$ Gaussian probability distribution $\Rightarrow$ all moments characterized by second-order moments $\Rightarrow$ state completely characterized by covariance matrix. In general, the Isserlis theorem applies to distributions with $\langle X_i X_j\rangle=\langle X_j X_i\rangle$, so the symmetrized version of the covariance matrix $\boldsymbol{\sigma}$ is necessary here.