Answer: No, it is not possible.

Let $T$ be an anti-Pascal pyramid with $n$ rows, containing every integer from 1 to $1+2+\cdots+n$, and let $a_{1}$ be the topmost number in $T$ (Figure 1). The two numbers below $a_{1}$ are some $a_{2}$ and $b_{2}=a_{1}+a_{2}$, the two numbers below $b_{2}$ are some $a_{3}$ and $b_{3}=a_{1}+a_{2}+a_{3}$, and so on and so forth all the way down to the bottom row, where some $a_{n}$ and $b_{n}=a_{1}+a_{2}+\cdots+a_{n}$ are the two neighbours below $b_{n-1}=a_{1}+a_{2}+\cdots+a_{n-1}$. Since the $a_{k}$ are $n$ pairwise distinct positive integers whose sum does not exceed the largest number in $T$, which is $1+2+\cdots+n$, it follows that they form a permutation of $1,2, \ldots, n$.

Figure 1

Figure 2

Consider now (Figure 2) the two 'equilateral' subtriangles of $T$ whose bottom rows contain the numbers to the left, respectively right, of the pair $a_{n}, b_{n}$. (One of these subtriangles may very well be empty.) At least one of these subtriangles, say $T^{\prime}$, has side length $\ell \geqslant\lceil(n-2) / 2\rceil$. Since $T^{\prime}$ obeys the anti-Pascal rule, it contains $\ell$ pairwise distinct positive integers $a_{1}^{\prime}, a_{2}^{\prime}, \ldots, a_{\ell}^{\prime}$, where $a_{1}^{\prime}$ is at the apex, and $a_{k}^{\prime}$ and $b_{k}^{\prime}=a_{1}^{\prime}+a_{2}^{\prime}+\cdots+a_{k}^{\prime}$ are the two neighbours below $b_{k-1}^{\prime}$ for each $k=2,3 \ldots, \ell$. Since the $a_{k}$ all lie outside $T^{\prime}$, and they form a permutation of $1,2, \ldots, n$, the $a_{k}^{\prime}$ are all greater than $n$. Consequently,
$$\begin{array}{r} b_{\ell}^{\prime} \geqslant(n+1)+(n+2)+\cdots+(n+\ell)=\frac{\ell(2 n+\ell+1)}{2} \\ \geqslant \frac{1}{2} \cdot \frac{n-2}{2}\left(2 n+\frac{n-2}{2}+1\right)=\frac{5 n(n-2)}{8} \end{array}$$
which is greater than $1+2+\cdots+n=n(n+1) / 2$ for $n=2018$. A contradiction.