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My solution for $n=3$ (upon Suvrit's request):

To begin with, we solve $A+A^T=M$ together with $A{\bf1}={\bf1}$ (no inequality for the moment). This is a linear system in $A$, which consists in $9$ equations in $9$ unknowns. However, it is not Cramer, because the set of skew-symmetric matrices $B$ such that $B{\bf1}=0$ is one-dimensional, spanned by $$\begin{pmatrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \end{pmatrix}.$$ In particular, there is a condition for solvability in $A$, but this condition is met by the assumption that $\sum_{i,j}m_{ij}=6$. Notice that $a_{ii}=\frac12m_{ii}$.

Thus there is a solution $A$, and every solution is of the form $A+aB$. There remains to find $a$ so as to satisfy the inequality $A+aB\ge0_n$. For this, let us denote $\mu$ the lower bound of $(a_{12},a_{23},a_{31})$, and $\nu$ that of $(a_{21},a_{13},a_{32})$.

Claim: we have $\nu+\mu\ge0$. This inequality allow us to find an $a$ such that $a_{12}+a,a_{23}+a,a_{31}+a,a_{21}-a,a_{13}-a,a_{32}-a\ge0$, which solves the problem.

Proof of the claim: we have $a_{12}+a_{21}=m_{12}\ge0$, $a_{12}+a_{13}=1-\frac12m_{11}\ge0$ because of the assumption that $m_{ij}\le2$, and finally $$a_{12}+a_{32}=a_{12}+a_{21}+a_{32}+a_{23}-a_{21}-a_{23}=m_{12}+m_{23}+\frac12m_{22}-1.$$ Form the assumption, this is equal to $$2-m_{13}-\frac12(m_{11}+m_{33})\ge0.$$ Finally, every sum $a_{ij}+a_{ji}$, $a_{ij}+a_{ik}$ and $a_{ij}+a_{kj}$ of elements of both sets is non-negative, hence $\mu+\nu\ge0$. Q.E.D.

Adapting this proof to higher $n$ seems difficult, but not impossible. Let us define $$s_A(I,J)=\sum_{i\in I,j\in J}a_{ij}.$$ If $A$ is any solution of $A+A^T=M$ and $A{\bf1}={\bf1}$, where $M$ meets the assumptions above, then for every $I,J$, we have $$s_A(I,J^c)+s_A(J,I^c)=|I|+|J|-s_M(I,J)\ge0.$$ Likewise, $a_{ij}+a_{ji}=m_{ij}\ge0$ for every $i,j$.

We have therefore reduced our question to the following one

Suppose that a matrix $A\in M_n({\mathbb R})$ satisfies $a_{ij}+a_{ji}\ge0$ for every $i,j$, and $s_A(I,J^c)+s_A(J,I^c)\ge0$ for every index sets $I,J$. Is it true that there exists a skew-symmetric matrix $B$, satisfying $B{\bf1}={\bf0}$, such that $A+B$ is entrywise non-negative?

(Remark that for such $B$, one has $s_B(I,J^c)+s_B(J,I^c)\equiv0$.)

A side remark: this set of assumptions about $A$ is redundant. All of them derive from the smaller set of inequalities $$a_{ij}+a_{ji}\ge0,\quad\forall i,j,\qquad s_A(I,I^c)\ge0,\quad\forall I.$$ As a matter of fact, one has $$s_A(I,J^c)+s_A(J,I^c)=s_A(I\setminus J,I\setminus J)+s_A(J\setminus I,J\setminus I)+s_A(I\cap J,(I\cap J)^c)+s_A((I\cup J)^c,I\cup J)$$ and $s_A(K,K)\ge0$ follows from $a_{ij}+a_{ji}\ge0$.

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Adapting this proof to higher $n$ seems difficult, because the kernel of the linear system the set of skew-symmetric matrices $B$ such that but not impossible.Let us define$B{\bf1}=0$) has dimension $s_A(I,J)=\sum_{i\in I,j\in J}a_{ij}.$$If \frac12(n-1)(n-2). At least, the system formed by A is any solution of A+A^T=M and A{\bf1}={\bf1} still has a solution. The set of solutions is still of the form A{\bf1}={\bf1}, where A+aB, and its invariants are M meets the quantities a_{ij}+a_{ji}, assumptions above, then for every \sum_{j(\ne i)}a_{ij} and I,J, we have\sum_{i(\ne j)}a_{ij}. All of them are non-negatives_A(I,J^c)+s_A(J,I^c)=|I|+|J|-s_M(I,J)\ge0.$$Likewise, being respectively equal$a_{ij}+a_{ji}=m_{ij}\ge0$for every$i,j$. We have therefore reduced our question to the following one Suppose that a matrix$m_{ij}$, A\in M_n({\mathbb R})$ satisfies $1-\frac{1}{2}m_{ii}$ a_{ij}+a_{ji}\ge0$for every$i,j$, and $$\sum_{i(\ne j)}m_{ij}+\frac12m_{ii}-1=\frac12\left(\sum_{k,l}m_{kl}-\sum_{k,l(\ne j)}m_{kl}\right)-1\ge\frac12(2n-2(n-1))-1=0.$$The fact s_A(I,J^c)+s_A(J,I^c)\ge0$ for every index sets $I,J$. Is it true that these invariant quantities are non-negative is compatible with the requirement there exists a skew-symmetric matrix $B$, satisfying $B{\bf1}={\bf0}$, such that $A\ge0_n$, but it must not be sufficient, as we did not use all of the A+B$is entrywise non-negative? (necessary) assumptions.Remark that for such$B$, one has$s_B(I,J^c)+s_B(J,I^c)\equiv0$.) 4 added 193 characters in body My solution for$n=3$(upon Suvrit's request): To begin with, we solve$A+A^T=M$together with$A{\bf1}={\bf1}$(no inequality for the moment). This is a linear system in$A$, which consists in$9$equations in$9$unknowns. However, it is not Cramer, because the set of skew-symmetric matrices$B$such that$B{\bf1}=0$is one-dimensional, spanned by $$\begin{pmatrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \end{pmatrix}.$$ In particular, there is a condition for solvability in$A$, but this condition is met by the assumption that$\sum_{i,j}m_{ij}=6$. Notice that$a_{ii}=\frac12m_{ii}$. Thus there is a solution$A$, and every solution is of the form$A+aB$. There remains to find$a$so as to satisfy the inequality$A+aB\ge0_n$. For this, let us denote$\mu$the lower bound of$(a_{12},a_{23},a_{31})$, and$\nu$that of$(a_{21},a_{13},a_{32})$. Claim: we have$\nu+\mu\ge0$. This inequality allow us to find an$a$such that$a_{12}+a,a_{23}+a,a_{31}+a,a_{21}-a,a_{13}-a,a_{32}-a\ge0$, which solves the problem. Proof of the claim: we have$a_{12}+a_{21}=m_{12}\ge0$,$a_{12}+a_{13}=1-\frac12m_{11}\ge0$because of the assumption that$m_{ij}\le2$, and finally $$a_{12}+a_{32}=a_{12}+a_{21}+a_{32}+a_{23}-a_{21}-a_{23}=m_{12}+m_{23}+\frac12m_{22}-1.$$ Form the assumption, this is equal to $$2-m_{13}-\frac12(m_{11}+m_{33})\ge0.$$ Finally, every sum$a_{ij}+a_{ji}$,$a_{ij}+a_{ik}$and$a_{ij}+a_{kj}$of elements of both sets is non-negative, hence$\mu+\nu\ge0$. Q.E.D. Adapting this proof to higher$n$seems difficult, because the kernel of the linear system the set of skew-symmetric matrices$B$such that$B{\bf1}=0$) has dimension$\frac12(n-1)(n-2)$. At least, the system formed by$A+A^T=M$and$A{\bf1}={\bf1}$still has a solution. The set of solutions is still of the form$A+aB$, and its invariants are the quantities$a_{ij}+a_{ji}$,$\sum_{j(\ne i)}a_{ij}$and$\sum_{i(\ne j)}a_{ij}$. All of them are non-negative, being respectively equal to$m_{ij}$,$1-\frac{1}{2}m_{ii}$and $$\sum_{i(\ne j)}m_{ij}+\frac12m_{ii}-1=\frac12\left(\sum_{k,l}m_{kl}-\sum_{k,l(\ne j)}m_{kl}\right)-1\ge\frac12(2n-2(n-1))-1=0.$$ The fact that these invariant quantities are non-negative is compatible with the requirement that$A\ge0_n\$, but it must not be sufficient, as we did not use all of the (necessary) assumptions.

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