Solution of an equation and a system of inequalities Consider an integer $n \geq 1$, a positive real number $A$ and a collection of nonnegative real numbers $\{a_{i,j}\}$ defined for $(i,j) \in \{1,\cdots,n\}^2$.
I want to find necessary and sufficient conditions on $A$ and the $\{a_{i,j}\}$ for the following system to have a nonnegative solution $(x_1,\cdots,x_n)$:
\begin{equation}
\left\{
\begin{array}{ll}
         x_1 + \cdots + x_n = A & \\
         \forall 1 \leq i \leq j \leq n, x_i + \cdots + x_j \leq a_{i,j}
\end{array}
\right.
\end{equation}
A necessary condition is easily obtained: for every covering $[i_1=1,j_1] \cup [i_2,j_2] \cup \cdots \cup [i_m,j_m=n]$ of $\{1,\cdots,n\}$, the inequality $a_{i_1,j_1} + \cdots + a_{i_m,j_m} \geq A$ must hold. Notice that I do not assume that $a_{i,j'} \geq a_{ij}$ when $j>j'$, which explains why the condition considers coverings and not only partitions of $\{1,\cdots,n\}$.
Is it also sufficient? I believe so (I have checked when $n \leq 4$), but I was not able to prove it. I have tried to do it by induction and by applying one of the theorems of the alternative (affine Farkas lemma).
Thanks a lot in advance.
 A: It seems the following.
Theorem. The following conditions are equivalent: 
i. (The solvability of the system) There exists a sequence $(x_1,\dots, x_n)$ of non-negative numbers such that $x_1+\dots x_n=A$ and $x_i+\dots x_j\le a_{ij}$ for each $1\le i\le j\le n$.
ii. (The compatibility of the weighted coverings). For each collection $\{\lambda_{ij}:1\le i\le j\le n\}$ of non-negative real numbers and each non-negative real number $\lambda$ such that for each $1\le k\le n$  
$$\lambda\le \sum_{i\le k\le j}\lambda_{ij}$$ we have 
$$\sum \lambda_{ij} a_{ij}\ge\lambda A.$$ 
iii. (The compatibility of the coverings). For each family $\mathcal I$ of discrete segments of the form $[i, j]=\{i, i+1,\dots, j\}$ with $1\le i\le j\le n$ such that $\bigcup\mathcal I=[1,n]$ we have
$$\sum\{ a_{ij}:[i,j]\in\mathcal I\}\ge A.$$ 
Proof. 
$(i)\Rightarrow (ii)$. It is obvious. 
$(ii)\Rightarrow (i)$. We shall use the following 
Theorem [Cha, §37, Th.1]. A system $(\alpha_k, x)\le \beta_k$ ($k=1,\dots m$) of linear inequalities has a solution iff the equality $\sum_{k=1}^m \lambda_k\alpha_k=0$ with non-negative scalars $\lambda_k$ implies an inequality $\sum_{k=1}^m \lambda_k\beta_k \geq 0$.$\square$
We consider a solvability of the following system of linear inequalities. 
$(\alpha_{ij}, x)\le a_{ij}$, for $1\le i\le j\le n$ and the $k$-th coordinate of the vector $\alpha_{ij}$ is $1$ if $i\le k \le j$ and $0$, otherwise.
$-(\delta_{k}, x)\le 0$, for $1\le k\le n$ and the $k$-th coordinate of the vector $\delta_k$ is $1$, and the other coordinates are $0$. 
$-(e, x)\le A$, and the vector $e$ has all coordinates $1$.
Now we apply Theorem to this system. Assume that 
$$\sum \lambda_{ij}\alpha_{ij}-\sum \lambda_k\delta_k-\lambda =0$$
(all $\lambda$’s are non-negative). 
We need to check 
$$\sum \lambda_{ij}a_{ij}\ge\lambda A.$$ 
which is true by Condition ii, because for each $k$ we have 
$$\lambda\le \lambda+\lambda_k=\sum_{i\le k\le j}\lambda_{ij}.$$
$(ii)\Rightarrow (iii)$. Condition iii is just a simplified case of Condition ii, when $\lambda=1$ and all $\lambda_{ij}\in \{0,1\}$. 
$(iii)\Rightarrow (ii)$. We shall use the following 
Splitting Lemma. Let $n$ and $\lambda$ be positive integers, $\mathcal I$ be an indexed family of discrete segments of the form $[i, j]=\{i, i+1,\dots, j\}$ with $1\le i\le j\le n$ such that for each index $1\le k\le n$ there exist exactly $\lambda$ elements $I$ of the family $\mathcal I$ such that $I\ni k$. Then the family $\mathcal I$ 
can be split into $\lambda$ subfamilies $\mathcal I_1,\dots, \mathcal I_\lambda$ such that every family $\mathcal I_\kappa$ is just a partition of the discrete segment  $[1, n]$ into segments. 
Proof. It suffices to separate from the family $\mathcal I$ the first family $\mathcal I_1$ and then continue by a similar manner. As a first segment $I_1=[i_1,j_1]\in\mathcal I_1$ we take an arbitrary segment of the family $\mathcal I$ which begins from $1$. If $j_1=n$ the we are done, so we shall follow the opposite case. Consider a family $\mathcal I’$ of segments of the family $\mathcal I$, containing $j_1+1$. By Lemma’s conditions  $|\mathcal I’|=\lambda$. If there are no segment $I_2\in\mathcal I’$ which begins from $j_1+1$ then $j\in \{I_1\}\cup\mathcal I’$,  which contradicts Lemma’s conditions. So we add the segment $I_2=[i_2:= j_1+1, j_2]$ to the family $\mathcal I_1$. Continuing in such the way, at some step $k$ we obtain $j_k=n$, which witnesses that we have constructed a partition $\mathcal I_1$ of the segment $[1,n]$.$\square$
Now we assume that there exist a collection $\{\lambda_{ij}:1\le i\le j\le n\}$ of non-negative real numbers and a positive real number $\lambda$ such that for each $1\le k\le n$  
$$\lambda\le \sum_{i\le k\le j}\lambda_{ij}$$ but holds Bounding Inequality 
$$\sum \lambda_{ij} a_{ij}<\lambda A.$$ 
Substituting all $\lambda$’s by their sufficiently good approximations by non-negative rational numbers, without loss of generality we may assume that all $\lambda$’s are rational. Moreover, multiplying all $\lambda$’s by a  common multiple of their denominators, without loss of generality we may assume that all $\lambda$’s are non-negative integers. Let $\mathcal I$ be an indexed family of discrete segments of the form $[i, j]=\{i, i+1,\dots, j\}$ with $1\le i\le j\le n$ which contains a $\lambda_{ij}$ copies of a segment $[i, j]$ for every $1\le i\le j\le n$.  
Now we monotonize $\{a_{ij}\}$’s by putting $a_{ij}’=\min\{a_{i’j’}:i’\le i$, $j\le j’\}$. This substitution will keep Bounding Inequality $\sum \lambda_{ij} a_{ij}’<\lambda A.$ Also Bounding Inequality is kept by a substitution $\lambda\to\lambda’=\min\{|\mathcal I_k|:1\le k\le n\}$, where $\mathcal I_k=\{I\in\mathcal I:I\ni k\}$ for each  $1\le k\le n$. 
Now we start to cut off the segments of the family $\mathcal I$, while keeping both Bounding Inequality and the $\lambda’$.  Let $k$ be the smallest number such that $|\mathcal I_k|>\lambda’$. If there is no segment $I\in I_k$ such that $I$ begins from $k$ then  $k>1$ and $\mathcal I_{k-1}\supset \mathcal I_k$, so $|\mathcal I_{k-1}|>\lambda’$, which contradicts to the choice of $k$. So we take an arbitrary segment $I=[k,j]\in I_k$ which begins from $k$ and substitute it by a segment $[k+1,j]$ (which may be empty). This substitution keeps both Bounding Inequality (this follows from the monotonization, which we did earlier) and the $\lambda’$.  Continuing in such the way, at some step we obtain that there are no segments to cut. It means that $|\mathcal I_k|=\lambda’$ for each $k$, so the family $\mathcal I$ satisfies the condition of Splitting Lemma, so it can be spilt into $\lambda’$ subfamilies $\mathcal I_1,\dots, \mathcal I_{\lambda’}$ such that every family $\mathcal I_\kappa$ is a partition of the discrete segment  $[1, n]$ into segments.  By Condition iii,  
$$\sum\{ a_{ij}:[i,j]\in\mathcal I_\kappa\}\ge A,$$ 
for each family $\mathcal I_\kappa$. Summing up all these inequalities, we obtain a contradiction with Bounding Inequality.$\square$
References
[Cha] V. S. Charin, Linear transformations and convex sets, Kyiv: Vyshcha shkola, 1978. (in Russian).
