jmc

Let the $k$th number in the $n$th row be $a(n,k)$. Writing out some numbers, we find that $a(n,k)=2^{n-1}(n+2k-2)$.[1] We wish to find all $(n,k)$ such that $67|a(n,k)=2^{n-1}(n+2k-2)$. Since $2^{n-1}$ and $67$ are relatively prime, it follows that $67|n+2k-2$. Since every row has one less element than the previous row, $1\le k\le 51-n$ (the first row has $50$ elements, the second $49$, and so forth; so $k$ can range from $1$ to $50$ in the first row, and so forth). Hence $n+2k-2\le n+2(51-n)-2=100-n\le 100,$ it follows that $67|n-2k+2$ implies that $n-2k+2=67$ itself. Now, note that we need $n$ to be odd, and also that $n+2k-2=67\le 100-n⟹n\le 33$. We can check that all rows with odd $n$ satisfying $1\le n\le 33$ indeed contains one entry that is a multiple of $67$, and so the answer is $\frac{33+1}{2}=\boxed{17}$.