The problem numbered three (lol) from my one year ago blog post might be relevant, and I'm ~90% sure I've read this in Martin Gardner. EDIT: See this. Come on how such incredibly well known thing ended up in IMO eh ?

Kayak wrote: The problem from my one year ago blog post might be relevant, and I'm ~90% sure I've read this in Martin Gardner. EDIT: See this. I see you are right, It is indeed relevant.

Indeed: Chang, Hu, Lih and Shieh: "Exact Difference Triangles," Bulletin of the Institute of Mathematics, Academia Sinica, Volume 5, 1977, pp 191- 197, available at http://w3.math.sinica.edu.tw/bulletin/bulletin_old/d51/5120.pdf .

prague123 wrote: Indeed: Chang, Hu, Lih and Shieh: "Exact Difference Triangles," Bulletin of the Institute of Mathematics, Academia Sinica, Volume 5, 1977, pp 191- 197, available at http://w3.math.sinica.edu.tw/bulletin/bulletin_old/d51/5120.pdf . #exposed

I might have done a fakesolve but I proved that if n is a triangular number then 1 to n can be made antipascal triangle.

IMO2019 wrote: I might have done a fakesolve but I proved that if n is a triangular number then 1 to n can be made antipascal triangle. Just check your proof for $n=15$, or read the paper by the four Chinese authors in posting #4.

prague123 wrote: Just check your proof for $n=15$, or read the paper by the four Chinese authors in posting #4. $n = 15$ is possible, just read the paper('s page two) by the four Chinese authors in posting #4

Thank you at leasr I got some relief. My proof follows through functional approach of injectivity and surjectivity.

Kayak wrote: The problem numbered three (lol) from my one year ago blog post might be relevant, and I'm ~90% sure I've read this in Martin Gardner. EDIT: See this. Come on how such incredibly well known thing ended up in IMO eh ? Can we stop pretending IMO consists of "original" problems and feigning shock whenever it turns out somewhere somehow someone has already written about the problem? I mean I apologise if the last sentence was written in jest but this is a very strange cultural tendency on these forums and I think it should be discussed more.

It's great people point out where the problem was previously mentioned though that is I think very useful and should be done much more.

I'm not quite sure if "new" implies "original" or not, but the IMO regulation said that problems should be "new". 4.2 The proposals should, as far as possible, cover various fields of pre-university mathematics and be of varying degrees of difficulty. They should be new and may not have been suggested for or used in any other mathematics competition.

IMO 2015 P1 was also not original but that was a P1 and this is P3. Also @below aren't IMO P3/P6's supposed to be slightly less trivial compared to AMC 8 problems ?

Quoting something from a few months ago because I think it's relevant (granted this was for AMCs as opposed to the IMO but still): v_Enhance wrote: djmathman wrote: Of course, I suppose that's what test-solvers are for, and I'll admit I could have been a little more assertive in pointing out similarities to previous problems (I did this a lot in my analysis of the original longlist but not the second draft). David, I wouldn't worry about it. People love to talk about how everything is "super well-known" after the fact, regardless of whether it's past TSTST, formulas for counting Young tableaux, or Zimbabwe olympiads. Almost always these claims of how well-known things are end up being exaggarated (there are even cases of users accusing AMC of corruption because of a similar problem on an unrelated paid training program). My stance is that if you've seen the exact problem before, then good for you --- take your points, do your victory dance and move on. Let the other 95% of people who haven't seen it get a chance to see it too.

djmathman wrote: Quoting something from a few months ago because I think it's relevant (granted this was for AMCs as opposed to the IMO but still): In my opinion, this is on a totally different scale. An IMO problem, especially Problem 3, is supposed to be checked thoroughly by the Problem Selection Committee and the Jury for its originality. I think we all deserve an explanation of how this problem is allowed to appear on the IMO paper.

Maybe we can get over this and discuss the actual solution?

So, isn't there any normal solution?

I have posted a solution (its in greek, but I think it can be understandable) here

Google translated page attached.

Thank you green_dog_7983.

msinghal wrote: My solution from the competition, which I finally got around to writing up. It uses a global argument to show the statement for sufficiently large $n$. As originally written, $n=2018$ was not big enough for the solution to work, though with a minor fix (see the last paragraph of the solution) it does (barely) work.

not to dox but is this the infamous solution that got the highest score (mod 7) at the IMO???

Solved with prvocislo Latex and writeup tutorial by txc We define $\min(i)$ as the minimum term in row $i$, $\max(i)$ as the maximum term in row $i$. We call a number tiny if it is $\leq 2018$ and massive if it is $\geq 1009\cdot 2019-2018$. Claim: $\min(i)$ is tiny for $i=1,2,\dots,2018$. Proof: Note that $$\max(i) \leq \max(i+1) -\min(i+1)$$$$ \max(i+1)\geq \max(i) + \min(i+1)$$$$1+2+3+\dots+2018 \geq \max(2018) \geq \sum \min(i) \geq 1+2+3+\dots+2018$$which implies that equality must hold and as such our claim must be true. Claim: There are at most $2$ massive numbers in every row except possibly the bottom row. Proof: Note that the difference between two non-tiny numbers is at most $2019\times 1009-2019$ which is not massive. Each massive number must therefore be above a tiny number. Since there is only one tiny number per row, there are at most two massive numbers per row. Claim: There are at most $1010$ massive numbers in the bottom row. Proof: Note that two adjacent massive numbers forms a tiny number on top of the two massive numbers. However, note that the second last row only has one tiny number. Therefore, there can be only $1$ pair of adjacent massive numbers. Hence, we have at most $\frac{2018}{2}+1=1010$ massive numbers in the bottom row. Claim: If $x\geq i+(i+1)+...2018$, $x$ must be at most $i$ rows from the bottom row Proof: FTSOC suppose that $x$ is $k$ rows from the bottom, where $k\geq i+1$. Define $x_1,x_2,\dots,x_k$ such that $x=x1$ and $x{i+1}$ is the larger number below $x_i$ for $i=1,2,\dots,k-1$. Then \begin{align*} x_k &= x+(x_2-x_1)+(x_3-x_2)+\dots+(xk-x{k-1}) \\ &\geq \big(i+(i+1)+\dots+2018\big)+\big(1+2+3+\dots +(k-1)\big) \\ &\geq \big(i+(i+1)+\dots+2018\big)+\big(1+2+3+\dots +i\big) \\ &> 1+2+\dots+2018, \end{align*}a contradiction. Now any massive number $x\geq 46+47+\dots+2018$ must be at most 46 rows from the bottom. Combining the above claims, there are at most 1010 massive numbers in the bottom row and 2 massive numbers in the next 45 rows, totalling $1010+45\times 2<2018$ massive numbers, a contradiction. $\Box$

We claim that the answer is negative. Suppose we have an anti-Pascal triangle $T$ with $n$ rows, which contains every integer from $1$ to $x=1+2+...+n.$ We will prove that $n\leq 12$ which yields the conclusion. Consider sequences $(a_i),(b_i)$, where $a_1=b_1$ is the number at top of $T$ and $a_{i+1},b_{i+1}$ are the lower and the most numbers respectively immediately below $a_i$ for $i>0.$ [asy][asy] unitsize(1 cm); real x=1/sqrt(3); pair a1,a2,a3,b2,b3; a1=(0,2); a2=(-x,1); b2=(x,1); a3=(-2x,0); b3=(0,0); label("$a_1$",a1,N); label("$a_1+a_2$",a2,W); label("$a_2$",b2,E); label("$a_1+a_2+a_3$",a3,W); label("$a_3$",b3,E); label("$\dots$",(0,-1),S); draw((a1--a2),EndArrow); draw((a1--b2),EndArrow); draw((a2--a3),EndArrow); draw((a2--b3),EndArrow); draw((a3--(-3x,-1)),EndArrow); draw((a3--(-x,-1)),EndArrow); [/asy][/asy] Observe that $b_i=a_1+a_2+...+a_i,$ and since the most number in $T$ is $1+2+...+n$ it follows that $(a_i)$ forms a permutation of set $1,2,...,n.$ Thus in particular $b_n=x.$ [asy][asy] unitsize(1 cm); real h=sqrt(3); pair a,b,c; a=(-3,0); b=(3,0); c=(0,3*h); draw (a--b--c--a); draw ((0.5,0)--(-1.25,1.75*h)); draw ((1.5,0)--(2.25,0.75*h)); label("$a_n$",(0.5,0),NE); label("$b_n$",(1.5,0),NW); label("$T_1$",(-1.25,1.75*h/2)); label("$T_2$",(2.25,0.75*h/2)); [/asy][/asy] Partition $T$ onto anti-Pascals triangle $T_1,T_2$ as on picture (they includes all numbers of bottom row except $a_n,b_n$); clearly at least one of them (say $T_1$) has at least $m=\left\lceil \frac{n-2}{2}\right\rceil$ rows. But then it contains a sequence of numbers $(c_i)$ (which is arranged analogously to $(a_i)$) and so their sum $s\geq c_1+c_2+...+c_m.$ But if $a_i$ lies in $T_1$ then $a_{i+1}$ it does, and this is impossible since $a_n$ isn't included in $T_1,$ so $$2+3+...+n\geq s\geq (n+1)+(n+2)+...+(n+m)\implies n(n+1)-m(m+1)\geq 2mn +2.$$For even $n$ it gives $n^2-14n+8\leq0\implies n\leq 12$ and for odd $n$ gives $n^2-8n+7\leq 0\implies n\leq 7,$ so done. P.S. Thanks for noticing of typos below.

JAnatolGT_00 wrote: We claim that the answer is negative. Suppose we have an anti-Pascal triangle $T$ with $n$ rows, which contains every integer from $1$ to $x=1+2+...+n.$ We will prove that $n\leq 14$ which yields the conclusion. Consider sequences $(a_i),(b_i)$, where $a_1=b_1$ is the number at top of $T$ and $a_{i+1},b_{i+1}$ are the lower and the most numbers respectively immediately below $a_i$ for $i>0.$ [asy][asy] unitsize(1 cm); real x=1/sqrt(3); pair a1,a2,a3,b2,b3; a1=(0,2); a2=(-x,1); b2=(x,1); a3=(-2x,0); b3=(0,0); label("$a_1$",a1,N); label("$a_1+a_2$",a2,W); label("$a_2$",b2,E); label("$a_1+a_2+a_3$",a3,W); label("$a_3$",b3,E); label("$\dots$",(0,-1),S); draw((a1--a2),EndArrow); draw((a1--b2),EndArrow); draw((a2--a3),EndArrow); draw((a2--b3),EndArrow); draw((a3--(-3x,-1)),EndArrow); draw((a3--(-x,-1)),EndArrow); [/asy][/asy] Observe that $b_i=a_1+a_2+...+a_i,$ and since the most number in $T$ is $1+2+...+n$ it follows that $(a_i)$ forms a permutation of set $1,2,...,n.$ Thus in particular $b_n=x.$ [asy][asy] unitsize(1 cm); real h=sqrt(3); pair a,b,c; a=(-3,0); b=(3,0); c=(0,3*h); draw (a--b--c--a); draw ((0.5,0)--(-1.25,1.75*h)); draw ((1.5,0)--(2.25,0.75*h)); label("$a_n$",(0.5,0),NE); label("$b_n$",(1.5,0),NW); label("$T_1$",(-1.25,1.75*h/2)); label("$T_2$",(2.25,0.75*h/2)); [/asy][/asy] Partition $T$ onto anti-Pascals triangle $T_1,T_2$ as on picture (they includes all numbers of bottom row except $a_n,b_n$); clearly at least one of them (say $T_1$) has at least $m=\left\lceil \frac{n-2}{2}\right\rceil$ rows. But then it contains a sequence of numbers $(c_i)$ as we defined in the beginning, and so their sum $s\geq c_1+c_2+...+c_m.$ But if $a_i$ lies in $T_1$ then $a_{i+1}$ it does, and this is impossible since $a_n$ isn't included in $T_1,$ so $$2+3+...+n\geq s\geq (n+1)+(n+2)+...+(n+m)\implies n(n+1)-m(m+1)\geq 2mn -2.$$For even $n$ it gives $n\leq 14$ and for odd $n$ gives $n\leq 5,$ so the proof is over. Should not it be $n<14$ (strict) when $n$ is even, and $n\leq 7$ when $n$ is odd?

hydo2332 wrote: Actually, can anyone prove that there is not an anti-pascal triangle with $7$ or $6$ rows? If $n=6$, then it can be shown as follows. Instead of taking the absolute difference, take the sum, i.e. except for the numbers in the bottom row, each number is the sum of the two numbers immediately below it. By taking so nothing changes in the modulo 2: $|a-b|=a+b (mod 2)$. If we let the bottom numbers to be $a, b, c, d, e, f$, then the numbers of the 5th row would be $a+b, b+c, c+d, d+e, e+f$, 4th row $a+c, b+d, c+e, d+f$, 3rd row $a+b+c+d, b+c+d+e, c+d+e+f$, 2nd row $a+e, b+f$, and finally 1st row $a+b+e+f$ (Sum is done modulo $2$). After all, if add the all $\frac{6*7}{2}=21$ numbers, we will get even sum whereas $1+2+...+21=231$ is odd. Contradiction.

user1729 wrote: Quite same solution in pages 2-3-4 in the pdf here. I wonder if the contestants who solved this problem in the IMO have heard/read about this article before.

No such anti-Pascal triangle exists. Suppose there exists anti-Pascal triangle with $2018$ rows containing every integer from $1$ to $N = 1 + 2 + \cdots + 2018$. Define an apex-path to be a sequence of $2018$ adjacent integers in the anti-Pascal triangle, with no two integers being in the same row and the first term being the topmost integer of the anti-Pascal triangle. There exists a unique strictly increasing apex-path $x_1, x_2, \dots, x_{2018}$. Note that for $2 \leq i \leq 2018$, \[y_i = x_i - x_{i-1}\]are the numbers in distinct cells of the triangular array. Since \[x_{2018} = x_1 + y_2 + \cdots + y_{2018} \geq 1 + 2 + \cdots + 2018,\]$x_{2018} = N$. Cut off two smaller anti-Pascal triangles of maximum size from the left and right corners so that $1, 2, \dots, 2018$ are not included in either one. Let the left one have $A$ rows and the right one have $B$ rows. Then $A + B \geq 2016$. WLOG, $A \geq 1008$. The anti-Pascal triangle with $A$ rows has a unique strictly increasing apex-path $a_1, a_2, \dots, a_{A}$. Defining $b_i = a_i - a_{i-1}$ for $2 \leq i \leq A$, we have \[a_A = a_1 + b_2 + \cdots + b_A \geq 2019 + 2020 + \cdots + (2018 + A) \geq 2019 + 2020 + \cdots + 3026 > N.\]Contradiction.

Replace $2018$ with $2k$. We show this is true for all $k\ge 7$. Suppose for some $k\ge 7$ there existed an anti pascal triangle with side length $2k$ (side length just means the length of the bottom row) and numbers $1, 2, \ldots, 1 + 2 + \cdots + 2k$. Let $S = 1 + 2 + \cdots + 2k$. Draw an arrow from each number on the triangle to the larger of the two numbers immediately below it. Define the main chain of an antipascal triangle to be the chain of arrows from the top cell to the bottom row. Notice in the main chain, the differences between consecutive elements are pairwise distinct and are all on the triangle. The bottom number in the main chain is equal to the sum of all differences plus the top number. Since these numbers are all distinct, and there are $2k$ of them, the bottom of the chain must be at least $1 + 2 + \cdots + 2k$, so it is equal to $S$. In fact, this implies one of $1,2,\ldots, 2k$ is at the top of the main chain, and the others are at most one element to the left or right of the main chain because they are equal to the differences between consecutive elements in the chain. Color cells white if the chain of arrows starting from that cell does not end up at $S$, and black otherwise. All numbers $1, 2, \ldots, 2k$ must in a black cell themself or directly adjacent to one. Now WLOG assume that $S$ is on the right half of the bottom row. Then consider the unique $k\times k$ anti pascal triangle that has a bottom row coinciding with the $k$ leftmost cells of the original antipascal triangle. Notice that this anti pascal triangle has no black cells, because any chain from a number in this triangle cannot go outside of this triangle, and $S$ is not in this triangle. Now consider the unique $(k-1) \times (k-1)$ anti pascal triangle inside it that has a bottom row coinciding with the $k-1$ leftmost cells of the original triangle. Every number in this triangle is only adjacent to numbers in the $k\times k$ anti pascal triangle which are all white cells. Therefore, this triangle with side length $k-1$ cannot have any numbers between $1$ and $2k$, inclusive. Now consider the main chain of this triangle. The number on the bottom of this chain is the sum of at least $k-1$ elements in the chain. (Using the fact that $5(k-1) \ge 4k+2$). This sum is at least \[ (2k+1) + (2k+2) + \cdots + (3k-1) = \frac{5k(k-1)}{2} \ge \frac{k}{2} \cdot (4k+2) =S, \]which means that the bottom of this chain must also be equal to $S$. This is absurd since it must be a white cell.

The answer is no. Assume for contradiction that such a triangle exists. Consider an anti-Pascal triangle $T$ with $n$ rows which every integer from $1$ to $N := 1 + 2 + 3 + \ldots + n$, and construct its chain $\mathcal{C}_T$ as follows: start with the top element; for each $x$ in the current chain and not at the bottom of $T$, the next element is the maximum of its children. Let $\mathcal{C}_T$ be given by the sequence $a_1, \ldots, a_n$. By induction, we have $a_i \ge \tbinom{i+1}{2}$ for each $i$, and in particular $a_n = N$. Remark: The above inequality is the anti-Pascal analogue of the Hockey-stick identity in Pascal's triangle. Now consider two subtriangles $T_1$ and $T_2$ of $T$ with maximal total side length with the property that all three triangles share a common base, $T_1$ and $T_2$ are disjoint, and $N$ and its two neighbors are included in neither triangle. Note that both triangles do not contain any of the numbers $1, 2, \ldots, n$. Let $T_1$ have $t_1$ rows and $T_2$ have $t_2$ rows. WLOG $t_1 \ge \frac{n}{2}-1$. Let $\mathcal{C}_{T_1}=\{\ell_1, \ell_2, \ldots, \ell_{t_1}\}$. Then if $\Delta \ell_i = \ell_i-\ell_{i-1}$, \begin{align*} \ell_{t_1} &= \ell_1+[\Delta \ell_2+\ldots+\Delta \ell_{t_1}] \\ &\ge (n+1)+(n+2)+\ldots+(n+t_1) \\ &\ge (n+1)+(n+2)+\ldots+(n+n/2-1) \\ &> 1+2+\ldots+n = N, \end{align*}a contradiction.

Not too bad for $45$ mohs I think. I claim the answer is no. Suppose otherwise. Note that if we have an $A$ above a $B$ and $C$, then either $C=A+B$ or $B=A+C$. Let the top number be $a$. Then each row we add a number to it that is distinct. If $a$ and the numbers we add are not exactly $1 \cdots 2018$, then they add to something that is bigger than $1+2+\cdots+2018$. If we follow the path of these increases, we find that all numbers between $1$ and $2018$ are at most $1$ space away from a path from the top of the triangle to the bottom that only goes down-left or down-right. Now consider the two $1008$ length equilateral triangles positioned on the bottom left and bottom right equilateral triangle. Then one of these triangles has no numbers between $1$ and $2018$. This means that, using the a similar argument as before, that the bottom row of the triangle has a number which is at least $2019+2020+\cdots+3026$ (the smallest $1008$ numbers bigger than $2018$), which we can easily compute is bigger than $1+2+\cdots+2018$, contradiction.

For any given anti-Pascal triangle $T$ we prove the following: if there are $m$ rows, and the $m$ smallest numbers in the triangle are $a_1,a_2,\dots,a_m$, then there is a number that is at least $a_1+a_2+\dots+a_m$. Let the number in the top row be $b_1$. Note that if the smallest number in the $k$th row is $b_k$ then the if there is a number in the $k-1$th row that is $S$ then there is a number in the $k$th row that is at least $S+b_k$. Therefore, by induction there is a number in row $k$ that is at least $b_1+\dots+b_k$. For $k=m$, there is a number in the last row that is at least $b_1+\dots+b_m\ge a_1+\dots+a_m$ as desired. $~$ Now, suppose there were an anti-Pascal triangle with values $1$ to $1+2+\dots+2018=N$ then equality must hold for the relation \[b_1+\dots+b_m\ge a_1+\dots+a_m\]That is, each row contains exactly one of the numbers $\{1,2,\dots,2018\}$. Furthermore, in each row if we take the number $b_1, b_1+b_2, \dots, b_1+\dots+b_{2018}=N$ it forms a ninja path (as described in IMO 2023/5), and any of the numbers $\{1,2,\dots,2018\}$ must be adjacent to it. Thus, if we construct two more anti-Pascal triangles $T_1$ and $T_2$ who share the same bottom edge as our size-$2018$ one, such that $T_1$ includes the left-most number of $T$ and also the number two to the left of $N$, and $T_2$ includes the right-most number of $T$ and the number two to the right of $N$, then neither of them includes the numbers $\{1,2,\dots,2018\}$. WLOG let $T_1$ have at least $1008$ rows. Then, there is a number in $T_1$ that is at least \[2019+2020+\dots+3026>N\]contradiction.

The answer is no. In the triangle we can draw a path starting with the top number such that each number on the path is connected to the larger of the two numbers below it. Then the path must end in the bottom row on the largest number in the triangle. Each number on the path is the sum of an adjacent number and the previous number on the path. Therefore, the last number on the path is the sum of $2018$ other numbers on the triangle, so the numbers $1, 2, \dots 2018$ must be on or adjacent to the path. We can find another anti-Pascal triangle with $2018/2 -1 = 1008$ rows inside the original triangle such that none of the numbers in the new triangle is adjacent to the path. Then we can similarly draw another path in the new triangle. The maximum element of the triangle is at least $2019 + 2020 + \dots (2019 + 1007) > 1 + 2 + \dots 2018,$ contradiction. Therefore, the answer is no.

holy smokes this is exactly identical to @above (numbers and everything skull) repost from blog WHATS up guys welcome back to another episode of minecraft bedwars techno never dies!! anyway i i i An anti-Pascal triangle is an equilateral triangular array of numbers such that, except for the numbers in the bottom row, each number is the absolute value of the difference of the two numbers immediately below it. For example, the following is an anti-Pascal triangle with four rows which contains every integer from $1$ to $10$. \[\begin{array}{ c@{\hspace{4pt}}c@{\hspace{4pt}} c@{\hspace{4pt}}c@{\hspace{2pt}}c@{\hspace{2pt}}c@{\hspace{4pt}}c } \vspace{4pt} & & & 4 & & & \\\vspace{4pt} & & 2 & & 6 & & \\\vspace{4pt} & 5 & & 7 & & 1 & \\\vspace{4pt} 8 & & 3 & & 10 & & 9 \\\vspace{4pt} \end{array}\]Does there exist an anti-Pascal triangle with $2018$ rows which contains every integer from $1$ to $1 + 2 + 3 + \dots + 2018$? wait im very glad that actually copied over. also i had read about this at some point so i had an inkling of a solution. constructing the graph can also be motivated however by actually trying to construct an anti-pascal triangle and realizing that a big issue is knowing which of two adjacent vertices is bigger. in essence i just need to like not have a defeatist attitude try things as long as theres something to try + i really hope this is not a fakesolve The answer is no. Suppose otherwise. Let $N=1+2+\dots+2018$ and let $n=2018$. Draw a graph in the same structure as the original triangle, each number is a node. For any node $c$ not in the bottom row and the two nodes $a,b$ immediately below, we draw an edge between $c$ and the larger of $a$ and $b$. Since $c<\max\{a,b\}$ then clearly any top-to-bottom sequence is increasing. Specifically we'll take the "longest" top-to-bottom sequence, which goes from the topmost vertex $v_1\to v_2\to \dots \to v_n$ on the bottom row. Evidently there are also vertices $a_2,\dots,a_n$ directly adjacent to the $v_i$, where we have \[v_i+a_i=v_{i-1}\]for $i\ge 2$. (It helps very much to scout small cases or even better use the diagram provided as a visual aid!) Claim: The multisets \[\{v_1\}\cup \{a_2,\dots,a_n\}\]and \[\{1,2,\dots,n\}\]are equivalent. Proof: This claim is extracted by the fact that $a_2+\dots+a_n$ is "too big" most of the time and the rest of the cases are well-behaved. Specifically if $v_1\ge n+1$ then \[1+N\le 1+\dots+(n-1)+v_1\le a_2+\dots+a_n+v_1=v_n\le N\]a contradiction. If $v_1\le n$ we repeat this process, except now \[N=1+\dots+n\le a_2+\dots+a_n+v_1=v_n\le N\]so equality should hold giving the claim. The idea now is to try and find another top-to-bottom sequence completely outside of our first path (the corresponding $a_i$ should be disjoint as well!). If this holds, we know that the differences between adjacent vertices in the path is at least $n+1$. To have the corresponding $a_i$ also be disjoint, we essentially need the new path to not have any adjacent $v_i$ with the old path. Time for a terrible explanation. We should be able to find some completely valid sub-triangle of size at least $1008$ by PHP (either on the left or the right side of the first path) such that if the path remains within this sub-triangle we are fine. (Proof: Consider the first path's largest vertex, and its index on the bottom row.) There is clearly a path from the topmost vertex $a$ of the subtriangle to a vertex $b$ on the bottom row. We should make at least $1007$ additions. Since each addition is more than $n$, we make a total addition of at least \[(n+1)+\dots+(n+1007)=2539654>2037171=N\]a contradiction. (I remember seeing a proof that all $n\ge 7$ fail, this is probably just stronger bounding than what is above. Above probably works for something like $n\ge 15$, idk)

I claim the answer is no. Start at the top and take a path that goes to the bottom where we choose the larger of the two numbers below us. This must end at $N = 1 + 2 + \dots + 2018$ since the number at the bottom is the sum of $2018$ distinct terms in the pyramid which by size reasons implies it has to be $N$. So therefore, the number at the top and the numbers adjacent to the path must be the terms $1, 2, \dots, 2018$. Now consider the two paths starting from the leftmost and rightmost number $1011$th row. Note that the original big path cannot share a number with both these paths. Therefore one path must have its bottom number be at least $2019 + 2020 + \dots + 3026 = 2542680 > 2037171 = 1+ 2+ \dots + 2018$ so we are done.

Solved with Narwhal234. The answer is definitely not. Fix $n = 2018$. Consider a path starting from the topmost value, going downwards towards the larger value. As we move down this path, color the numbers we cross green. For each green number, color the other cell directly beneath it purple. We call this the grapevine. $\textbf{Claim.}$ The integers $1$ through $n$ appear exactly once in the grapevine. $\textit{Proof.}$ We add the purple numbers to the number up top to get the number at the bottom of the grapevine. This sum is at least $1 + 2 + \dots + n$, but this is the maximum value of the triangle. It follows that equality must hold, i.e. the grapes must consist of a permutation of $1$ through $n$ (including the top number.) Now consider the bottom row, split into three pieces: pre-grapevine, grapevine, post-grapevine. One of these three pieces has length at least $1008$, so there exists a subtriangle of size $1008$. We consider a grapevine in this subtriangle. The sum of the grapes, including the top number, is at least \[ 2019 + 2020 + \dots + (2018 + 1008) = 2018 \cdot 1008 + \frac{1008 \cdot 1009}{2} = 1009(2016 + 504) = 1009(2520) > \frac{2018 \cdot 2019}{2}. \]This is too big. Thus unfortunately we can make no wine, as desired.