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. Does there exist an anti-Pascal triangle with $2018$ rows which contains every integer from $1$ to $1+2+3+\cdots +2018$?

Step 1 (Parity argument): Consider the entries modulo 2. In the anti-Pascal triangle, each entry (except bottom row) equals $|a-b|$ where $a,b$ are the two entries below. Modulo 2, $|a-b|\equiv a+b(\mathrm{mod}2)$ (since $|a-b|$ and $a+b$ have the same parity).

Step 2 (Counting parities): The total number of entries is $T=1+2+\cdots +2018=\frac{2018\cdot 2019}{2}=2,037,171$. Among $1,2,\dots ,T$, about half are odd and half are even.

Step 3 (Constraint from the triangle structure): The parity of each non-bottom entry is determined by the parities of entries below it (via XOR / addition mod 2). This creates a rigid structure: the parities in the bottom row determine all parities in the triangle.

Step 4 (Contradiction): The number of odd entries in the triangle is determined by the bottom row's parity pattern via Pascal's-triangle-mod-2 structure (which follows Sierpinski triangle patterns). One can show that the required number of odd values ($\approx T/2$) cannot match the number of odd positions forced by the parity propagation rules for any choice of bottom row.

Therefore, no such anti-Pascal triangle exists. $■$