Lemma: Let $T$ be a stable triangular array with row sums $s_1, s_2, \ldots s_n$. Then there is a unique index $i$ such $a_{i,n}$ in $T$ can be increased by one to get a new stable triangular array $T^+$ with row sums $s_1, s_2, \ldots s_n + 1$. Additionally, if $s_n > s_{n-1}$, there is also a unique index $j$ such that $a_{j,n}$ in $T$ can be decreased by one to get a new stable triangular array $T^-$ with row sums $s_1, s_2, \ldots s_n - 1$, and these increases and decreases invert each other Proof: For $0 \leq i < j \leq n-1$ let $x_{i,j} = a_{i,n} - a_{i,j} - a_{j,n}$, and let $x_{j,i} = 1 - x_{i,j}$. Notice that all the $x_{i,j}$ are either $0$ or $1$ by the fact that $T$ is stable. Considering these as entries of an $n \times n$ matrix $M$ with $0$s down the diagonal, we get the adjacency matrix of a complete directed graph, a tournament. Note that if we increase $a_{i,n}$ by $1$ for some $i$, this is the same as increasing all $x_{i,k}$ by 1 for all $k \neq i$. This preserves the condition of stable arrays if and only if all $x_{i,k}$ are equal to $0$. Likewise, decreasing $a_{j,n}$ by $1$ requires all $x_{j,k}$ to be $1$. So we must find exactly one column of all 0s and exactly one column of all 1s in $M$, excluding the diagonal entries. We now show the tournament $M$ is transitive. If for $i < j < k$ we have $x_{i,j} = x_{j,k} = 0$, then \[0 = x_{i,j} + x_{j,k} = a_{i,n} - a_{i,j} - a_{j,k} - a_{k,n} \geq a_{i,n} - a_{i,k} - a_{k,n} = x_{i,k},\] so $x_{i,k} = 0$. Similarly, if $x_{i,j} = x_{j,k} = 1$, then \[2 = x_{i,j} + x_{j,k} = a_{i,n} - a_{i,j} - a_{j,k} - a_{k,n} \leq a_{i,n} + 1 - a_{i,k} - a_{k,n} = x_{i,k} + 1,\] so $x_{i,k} = 1$. Now if the tournament $M$ is not transitive, there exists a directed 3-cycle $i,j,k$. WLOG $i < j < k$, then the results we just proved give a direct contradiction. Since this tournament is transitive, it is well-known there exists exactly one $i$ such that $x_{i,k} = 1$ for all $k \neq i$, and likewise exactly one $j$ such that $x_{j,k} = 0$ for all $k \neq j$. These are the desired columns. Finally, doing the increment on $a_{i,k}$ changes all numbers in that column from 1 to 0, so it becomes the $j$ for the new array, and vice-versa. Thus these two operations are inverses of each other. $\blacksquare$ To prove the problem, we go by induction on $n$. The base case $n = 1$ is clear, since $a_{0,1}$ must be equal to $s_1$. Suppose it is true for $n-1$, and consider an arbitrary nondecreasing sequence $s_1, s_2, \ldots s_n$. By the inductive hypothesis we can choose a unique $(n-1)$-row stable triangular array with entries $a_{i,j}$ for $0 \leq i < j \leq n-1$ that satisfy the row sums $s_1, s_2, \ldots s_{n-1}$. It remains to show there is a unique way to assign the elements of row $n$, namely $a_{i,n}$ for $0 \leq i \leq n-1$, to satisfy the row sum $s_n$ while still keeping the array stable. We now prove that we can uniquely assign row $n$ by another induction on $s_n$. The base case is $s_n = s_{n-1}$. Note that $a_{i,n-1} + a_{n-1,n} \leq a_{i,n}$. Summing this up for all $i$ from $0$ to $n-2$ gets that $s_{n-1} + (n-1)a_{n-1,n} \leq s_n - a_{n-1,n}.$ Since $s_n = s_{n-1}$ and $a_{n-1,n} \geq 0$, we must have equality, so the unique row $n$ that works is $a_{n-1,n} = 0$ and $a_{i,n} = a_{i,n-1}$. Assume the inductive hypothesis is true for $s_n$. Then we have a unique stable triangular array $T$ with row sums $s_1, s_2, \ldots s_n$. By the lemma, there is a unique way to increment an entry in row $n$ of $T$ to get a new stable triangular array $T^+$ with row sums $s_1, s_2, \ldots s_n + 1$, showing existence. To show uniqueness, suppose we had a second array $T_1^{+}$. Apply the lemma to show there is a unique way to decrement an entry in row $n$ to get a new stable triangular array $T_1$ with row sums $s_1, s_2, \ldots s_n$. By the inductive hypothesis, $T = T_1$. We can obtain $T_1^{+}$ by incrementing an entry of $T_1$, the same as doing the earlier decrement in reverse. We can also obtain $T^+$ by incrementing an entry of $T$. Finally, the statement of the lemma claims that there is a unique way to increment an entry of $T$ to get a new stable array, so the increments done to $T$ and $T_1$ must have been the same. This shows that $T_1^{+} = T^+$, so uniqueness of the stable array with row sums $s_1, s_2, \ldots s_n + 1$ has been shown. This completes both inductions and the proof.

As in the previous solution, we induct on $n$ and use the inductive hypothesis (IH) to determine the entries of the triangle for the first $n-1$ rows (pretty much any solution has to do this). Let $L = \sum_{j=1}^{n-1} a_{0,j}$. Observe that by summing the condition for $i = 0$, $k = n$, and $1 \leq j \leq n-1$, we get \[ L + s_n \leq na_{0,n} \leq L + s_n + (n-1). \] Therefore, \[ a_{0,n} = \left\lceil \frac{ L + s_n}{n} \right\rceil. \] Now for $i = 1, 2, \ldots, n-1$, define $s'_i = s_{i+1} - a_{0,i+1}$. Apply the IH for the sums $s'_1, s'_2, \ldots, s'_{n-1}$. This gives us all the terms of the array we want except those in the first column (i.e. all the terms $a_{0,i}$ with $1 \leq i \leq n$). There is also no discrepancy in the overlap with the previous application of the IH; this is by a third application of the IH for $n-2$ and the sums $s'_1, s'_2, \ldots, s'_{n-2}$. So by combining these uses of the IH we've determined every entry of the array. It remains to check the condition. By the IH it holds for all $(i,j,k)$ with either $i \neq 0$ or $k \neq n$. So all we have to check is that for each $j$, \[ a_{0,j} + a_{j,n} \leq a_{0,n} \leq a_{0,j} + a_{j,n} + 1. \] For $j < j'$, we claim $|(a_{0,j} + a_{j,n}) - (a_{0,j'} + a_{j',n})| \leq 1$. To show this, note the LHS is equal to \[ |(a_{j,n} - a_{j,j'} - a_{j',n}) - (a_{0,j'} - a_{0,j} - a_{j,j'})|. \] Both terms in parentheses are 0 or 1 by the given condition, and the claim follows. To finish, note that if for a given $j$ we have $a_{0,j} + a_{j,n} > a_{0,n}$, then for all $j'$ we have $a_{0,j'} + a_{j',n} \geq a_{0,n}$ by the claim above. Summing this up for all $j'$ contradicts $a_{0,n} = \left\lceil \frac{ L + s_n}{n} \right\rceil$. The proof for $a_{0,j} + a_{j,n} + 1 < a_{0,n}$ being impossible is the same.

Denote $\Phi$ to take either the value $0$ or $1$. Then we see that $a_{i,k}=a_{i,j}+a_{j,k}+\Phi$. We will prove by induction on rows that for any $n$ there exists a unique representation. Base case is trivial, just let $a_{0,1}=s_1$. Now suppose that rows $1\to n-1$ have already been filled in uniquely. We will prove by induction on the element number of row $n$ that row $n$ can also be filled in uniquely. Base case: We see that for $k=0\to n-2$, that $$a_{k,n}=a_{k, n-1}+a_{n-1, n}+\Phi$$ and so $$s_n=\sum_{k=0}^{n-1}a_{k, n}=na_{n-1,n}+(n-1)\Phi+\sum_{k=0}^{n-2}a_{k, n-1}=s_{n-1}+na_{n-1,n}+(n-1)\Phi$$ $$\implies s_n-s_{n-1}=na_{n-1,n}+(n-1)\Phi$$ But note that $(n-1)\Phi$ can take any integer value $\in[0, n-1]$ and $s_n-s_{n-1}$ is a constant so by the Division Lemma $a_{n-1,n}$ is unique. Now suppose that $a_{n-1,n}, a_{n-2, n}, \ldots , a_{n-i+1,n}$ are unique, and we want to prove $a_{n-i,n}$ is also unique. Note that $a_{k,n}=a_{k, n-i}+a_{n-i,n}+\Phi$ for $k=0\to n-i-1$. Thus, $$s_n=\sum_{k=0}^{n-i}a_{k,n} + \sum_{k=n-i+1}^{n-1}a_{k,n}=(n-i+1)a_{n-i,n}+\sum_{k=0}^{n-i-1}a_{k, n-i} +\sum_{k=n-i+1}^{n-1}a_{k,n} + (n-i)\Phi = s_{n-i}+\sum_{k=n-i+1}^{n-1}a_{k,n} +(n-i+1)a_{n-i,n} + (n-i)\Phi $$ $$\implies s_n-s_{n-i}-\sum_{k=n-i+1}^{n-1}a_{k,n} = (n-i+1)a_{n-i,n} + (n-i)\Phi$$ But note that $(n-i)\Phi$ can take the integer values $\in [0, n-i]$ and the LHS is a constant by the induction step so therefore $a_{n-i,n}$ is also unique, by the Division Lemma. Thus, our induction is complete, and we have prove that row $n$ is uniquely determined by rows $1\to n-1$. Thus, for any $n$ the sequence $s_1, s_2, \ldots , s_n$ uniquely determines a stable triangular array. $\Box$

I tried to construct an appropriate array and found myself flailing to understand where the $+1$s are gathered. Seeing that we certainly have $a_{i,j}\approx a_{i,i+1}+a_{i+1,i+2}+\dots +a_{j-1,j}$ while trying to find a way to guarantee $a_{i,k}-a_{i,j}-a_{j,k}\in\{0,1\}$ always, the known fact $\lfloor x+y\rfloor -\lfloor x \rfloor -\lfloor y\rfloor \in\{0,1\}$ for all real $x,y$ comes to mind. It follows that if we fix a sequence $x_1,x_2,\dots,x_n$ of non-negative reals and define $a_{i,j}=\left\lfloor \sum_{i<p\le j} x_p \right\rfloor$, then the resulting triangular array is stable. If $s_1\le \dots\le s_n$ are prescribed, we can choose $x_1,x_2,\dots$ recursively, with $x_{k+1}$ chosen in such a way that the increasing function $f:[0,\infty)\to [s_k,\infty)$, $$f(t)=\lfloor t+(x_1+\dots+x_k)\rfloor +\lfloor t+(x_1+\dots+x_{k-1})\rfloor +\dots +\lfloor t\rfloor$$has value $s_{k+1}$. (We can ensure that the floors step at pairwise different times if we take care to never let any $x_{i+1}+\dots+x_j$ be an integer.) This proves existence. To prove uniqueness, just induct. Row $1$ is clearly unique; suppose row $1,2,\dots,k-1$ is unique. Note that for each $m=k-1,k-2,\dots,1$, $$a_{0,k}+\dots+a_{m,k}=(a_{0,m}+a_{m,k}+[0/1])+(a_{1,m}+a_{m,k}+[0/1])+\dots+(a_{m-1,m}+a_{m,k}+[0/1])+a_{m,k}.$$If we know the value of the left-hand side, then by the induction hypothesis, we know the value of $(m+1)a_{m,k}$ plus a number between $0$ and $m$, from which the value of $a_{m,k}$ is obtained via division with remainder. This clearly allows us to inductively show that $a_{m,k}$ is uniquely defined for $m=k-1$, then for $m=k-2$, and so on, until $m=1$; then it's just $s_k$ minus the rest for $m=0$: unique as well. This shows that row $k$ is also unique, completing the induction.