On the positivity of matrices For given $n$, the following $n\times n$ real matrix $M=M^{T}$ is called positive, if 
$x^{T}M x\geq 0$
holds for all non-negative real $x_1,x_2,\cdots,x_n$,
where $x=(x_1,x_2,\cdots,x_n)^T$.
Notice here, the definition of positive is not the same as usual defination of positive semidefinite.
Now the goal is to characterize the set of all such $M$. In particular, I would like to show the following statement:
Such $M$ can always written as positive combination of some positive semidefinite matrix and some $E_{ij}$s, where $E_{ij}$ is the $n\times n$ matrix with the only non zero element 1 lies in the position $(i,j)$.
 A: Such matrices are called copositive in the literature. 
Moreover, the statement you want to show is known to be false. 
While I don't recall a counterexample right away,
an intuition for this is that it's NP-hard to check copositivity, but computationally easy to check the condition in your statement, via semidefinite programming, see e.g. this text by Pablo Parrilo.
EDIT: here is an explicit counterexample: the matrix
$$
M=\begin{pmatrix}
 1&-1& 1& 1&-1\\
-1& 1&-1& 1& 1\\
 1&-1& 1&-1& 1\\
 1& 1&-1& 1&-1\\
-1& 1& 1&-1& 1
\end{pmatrix}
$$
is copositive, but not equal to the sum of a positive semidefinite matrix $P$ and a nonnegative matrix $N$.
It is taken from an old paper by P.Diananda in 1962 (Proc. Cambridge Phil.Soc., vol 58(1962)), where it is also shown that $n=5$ is minimal size for which one has such a counterexample.
A quick way to see that $M\neq P+N$ is as follows. First of all notice that $N$ can be assumed to have 0s on the main diagonal. This the condition $M=P+N$, that we want to bring to a contradiction, is equivalent to the existence of a positive semidefinite $P$ with all 1s on the diagonal, and satisfying the conditions $M_{ij}-P_{ij}\geq 0$, for all $1\leq i<j\leq 5$. 
As $P$ is p.s.d., also for $i<j$ one has  $|P_{ij}|\leq 1$, as can be checked by computing
$x^\top P x$ for $x$ with $x_i=x_j=1$ and the remaining entries 0.  Thus $P_{ij}=-1$ for all $i,j$ s.t. $M_{ij}=-1$. Further, all $P$'s satisfying these condition form a convex set $\Sigma$. 
Thus if $P\in\Sigma$ then $\Pi P\Pi^\top\in\Sigma$, for any permutation matrix $\Pi$ commuting with $M$, as $M=\Pi M\Pi^\top=P+N=\Pi (P+N)\Pi^\top$, i.e. $\Pi (P+N)\Pi^\top$ is another representation of $M$. Hence, by an averaging argument over the group $G$ of permutation matrices commuting with $M$, we can select $P$ to commute with all such $\Pi$'s. (More explicitly, take $\frac{1}{|G|}\sum_{\Pi\in G}\Pi P\Pi^\top$.) It follows that there must exist $P$ of the form
$$
P=\begin{pmatrix}
 1&-1& p& p&-1\\
-1& 1&-1& p& p\\
 p&-1& 1&-1& p\\
 p& p&-1& 1&-1\\
-1& p& p&-1& 1
\end{pmatrix}, \quad -1\leq p\leq 1
$$
But none of such $P$'s are p.s.d., as its eigenvalues are
$$2 \, p - 1,
 -\frac{1}{2} \, p {\left(\sqrt{5} + 1\right)} - \frac{1}{2} \, \sqrt{5} + \frac{3}{2},
 \frac{1}{2} \, p {\left(\sqrt{5} - 1\right)} + \frac{1}{2} \, \sqrt{5} + \frac{3}{2}.
$$
