The "Delete" button was still active and I had the possibility to re-post my proof. I omitted to write a very important property. $a_j-a_i>0\Longrightarrow a_i<a_j,\;\forall i<j$. Results: $a_1<a_2<\dots<a_n$. Denote: $D_{i,j}=a_j-a_i,\; 1\le i<j\le n$; $d_i=D_{i,i+1},\; 1\le i<j\le n-1$. $\max_{1\le i<j\le n}D_{i,j}=D_{1,n}=\dfrac{n(n-1)}2$. $D_{1,n}=d_1+d_2+\dots+d_{n-1}\ge1+2+\dots+(n-1)=\dfrac{n(n-1)}2$. Results: $\{d_1,d_2,\dots,d_{n-1}\}=\{1,2,\dots,n-1\}$. And now, the property which I mentioned: the numbers $D_{i,j}$ are pairwise distinct. Denote $A=\{D_{i,j}|i,j\in\mathbb{N}, 1\le i<j\le n\}$; $B=\left\{1,2,\ldots,\dfrac{n(n-1)}2\right\}.$ The number of distinct pairs $(i,j)$ is $C_n^2=\dfrac{n(n-1)}{2}$. Results: $|A|\le\dfrac{n(n-1)}{2}=|B|$. From the initial condition $|A|=|B|$ and this equality occurs only if the numbers $D_{i,j}$ are pairwise distinct. Conclusion: $\forall k\in B$, exists exactly a pair $(i,j)$ such that $D_{i,j}=k$. $n\in\{2,3,4\}$ satisfy the condition: For $n=2: d_1=1$; For $n=3: d_1=1;d_2=2$; For $n=4: d_1=1;d_2=3;d_3=2$. For $n\ge5: \dfrac{n(n-1)}2\ge10$. a) $\exists 1\le i<j\le n$ such that $D_{i,j}=\dfrac{n(n-1)}2-1$. This fact is possible only if $d_1=1$ or $d_{n-1}=1$. WLOG, we can consider $d_1=1$ (reversing the order of $(d_1,d_2,\dots,d_{n-1})$, the requested property is satisfied). b) $\exists 1\le i<j\le n$ such that $D_{i,j}=\dfrac{n(n-1)}2-2$. This fact is possible only if $d_1=1$ and $d_{n-1}=2$. c) $\exists 1\le i<j\le n$ such that $D_{i,j}=\dfrac{n(n-1)}2-3>n-1$. This fact is possible in $2$ situations: c.1) the differences $d_1=1$ and $d_j=2$ are consecutive, hence $j=2\Longrightarrow n=3$, contradiction. c.2) the difference $d_i=3$ is the first or the last in the sequence $(d_1,d_2,\dots,d_{n-1})$, contradiction with the paragraph b). Final conclusion: $n\in\{2,3,4\}$.