Which matrices are covariances matrices? Let $V$ be a matrix. 
What conditions should we require so that we can find a random vector $X = (X_1, \dots, X_n)$ so that $V = Var(X)$? 
Of course necessary conditions are:


*

*All the elements on the diagonal should be positive

*The matrix has to be symmetric

*$v_{ij} \le \sqrt{v_{ii}v_{jj}}$ (Because of $Cov(X_i, X_j) \le \sqrt{Var(X_i) Var(X_j)})$


But I am sure these are not sufficient as I have a counterexample.
So what other properties we should require on a matrix so that it can be considered a covariance matrix? 
 A: I think I cleared this up sufficiently.
Okay, so 
1) If $V$ is not semi definite positive, then such a vector $X$ does not exists. (Since all covariances matrix are semi definite positive)
2) If $V$ is symmetric semidefinite positive, then such an $X$ exists! [0]
This implies that 
$$\text{exists a random vector X: V = Cov(X)} \iff \text{V is symmetric positive semidefinite}$$
Since we know that those I listed in the question are necessary condition for $V$, we deduce that all symmetric semidefinite positive matrices have elements on the diagonal $\ge 0$ and are such that $v_{ij} \le \sqrt{v_{ii}v_{jj}}$.
These are not sufficient though for a matrix to be semidefinite positive but sufficient conditions are well known, after all.

[0] Proof
Since $V$ is symetric is possible to find an orthogonal matrix $Q$ such that $V = QDQ^T$, where $D$ is a diagonal matrix whose values are the eigenvalues of $V$.
If $V$ is semipositive definite the elements of $D$ are all $\ge 0$, hence we can find $X$ such that $D = Cov(X)$ (just take all the variables independent with the specified variance)
It follows that the random vector $QX$ has covariance equal to $$Cov(QX) = QCov(X)Q^T = QDQ^T = V$$
