Integers in a triangle, and differences I read about the following puzzle thirty-five years ago or so, and I still do not know the answer.
One gives an integer $n\ge1$ and asks to place the integers $1,2,\ldots,N=\frac{n(n+1)}{2}$ in a triangle according to the following rules. Each integer is used exactly once. There are $n$ integers on the first row, $n-1$ on the second one, ... and finally one in the $n$th row (the last one). The integers of the $j$th row are placed below the middle of intervals of the $(j-1)$th row. Finally, when $a$ and $b$ are neighbours in the $(j-1)$th row, and $c$ lies in $j$-th row, below the middle of $(a,b)$ (I say that $a$ and $b$ dominate $c$), then $c=|b-a|$.
Here is an example, with $n=4$.
$$\begin{matrix} 6 & & 10 & & 1 & & 8 \\\\ & 4 & & 9 & & 7 \\\\ & & 5 & & 2 & & \\\\ & & & 3 & & & \end{matrix}$$
Does every know about this ? Is it related to something classical in mathematics ? Maybe eigenvalues of Hermitian matrices and their principal submatrices.
If I remember well, the author claimed that there are solutions for $n=1,2,3,4,5$, but not for $6$, and the existence was an open question when $n\ge7$. Can anyone confirm this ?
Trying to solve this problem, I soon was able to prove the following.

If a solution exists, then among the numbers $1,\ldots,n$, exactly one lies on each line, which is obviously the smallest in the line. In addition, the smallest of a line is a neighbour of the highest, and they dominate the highest of the next line.

The article perhaps appeared in the Revue du Palais de la Découverte. 
Edit. Thanks to G. Myerson's answer, we know that these objects are called Exact difference triangles in the literature. 
 A: This is the first problem in Chapter 9 of Martin Gardner, Penrose Tiles to Trapdoor Ciphers. In the addendum to the chapter, he writes that Herbert Taylor has proved it can't be done for $n\gt5$. Unfortunately, he gives no reference.
There may be something about the problem in Solomon W Golomb and Herbert Taylor, Cyclic projective planes, perfect circular rulers, and good spanning rulers, in Sequences and their applications (Bergen, 2001), 166–181,
Discrete Math. Theor. Comput. Sci. (Lond.), Springer, London, 2002, MR1916130 (2003f:51016).
See also http://www.research.ibm.com/people/s/shearer/dts.html and the literature on difference matrices and difference triangles.
EDIT. Reading a little farther into the Gardner essay, I see he writes,
The only published proof known to me that the conjecture is true is given by G. J. Chang, M. C. Hu, K. W. Lih and T. C. Shieh in "Exact Difference Triangles," Bulletin of the Institute of Mathematics, Academia Sinica, Taipei, Taiwan (vol. 5, June 1977, pages
191- 197).
This paper can be found at http://w3.math.sinica.edu.tw/bulletin/bulletin_old/d51/5120.pdf and the review is MR0491218 (58 #10483).
EDIT 2023: Brian Chen, YunHao Fu, Andy Liu, George Sicherman, Herbert Taylor, Po-Sheng Wu, Triangles of Absolute Differences, Chapter 11 (pages 115-124) in Plambeck and Rokicki, eds., Barrycades and Septoku: Papers in Honor of Martin Gardner and Tom Rodgers, MAA Press 2020, also gives a proof.
A: Not an answer, but a question about constraints mod 2. There you are just taking differences, and the number 0s and 1s must be the same for n that is 0 or 3 mod 4 (one different in the other two cases). On the face of it the left edge of the triangle could be any binary sequence. Does this give anything?
A: From what I remember there is now an actual proof by 6 authors (one is Andy Liu); I think the paper was presented at g4g9. The paper might have not been published yet.
It is my understanding that Herbert Taylor was true but way to complicated, I think that Liu & et  used his basic idea but simplified a lot the proof, which is now at an elementary level.
A: For small values of $n$, there is a relatively small state space to search.
In the most naive way possible, I found the following (showing the top row of triangle only):
1: 1 way: [1]
2: 4 ways: [1,3], [2,3], [3,1], [3,2]
3: 8 ways: [1,6,4], [2,6,5], [4,1,6], [4,6,1], [5,2,6], [5,6,2], [6,1,4], [6,2,5]
4: 8 ways: [6,1,10,8], [6,10,1,8], [8,1,10,6], [8,3,10,9], [8,10,1,6], [8,10,3,9] [9,3,10,8], [9,10,3,8]
5: 2 ways: [6,14,15,3,13], [13,3,15,14,6]
6: no ways

In particular, it is possible for $n = 5$, but not possible for $n = 6$.
The computation for $n = 7$ seems entirely feasible, and I'm happy to carry it out.
A: So for n=6, we know we cannot achieve the same thing, but perhaps this is the next best top row?: [6,20,22,3,21,13].  We could say this difference triangle has ``excess one''. What is the smallest excess for n>6?
