By induction, the $n$th entry of row $r$ is $2^{r-1}(r + 2n - 2):$ row $1$ gives $2^0(2n - 1),$ and summing two adjacent entries of row $r$ gives $2^{r-1}(r + 2n - 2)$ $+ 2^{r-1}(r + 2n)$ $= 2^r\bigl((r + 1) + 2n - 2\bigr),$ the formula for row $r + 1.$ Row $r$ has $51 - r$ entries, so $1 \le n \le 51 - r.$

Since $67$ is odd, an entry is a multiple of $67$ exactly when $67 \mid r + 2n - 2.$ As $n$ runs through row $r,$ the quantity $r + 2n - 2$ takes the values $r, r + 2, \ldots, 100 - r,$ all with the same parity as $r$ and all less than $134.$ So the only possible multiple of $67$ is $67$ itself, which requires $r$ odd and $r \le 67 \le 100 - r,$ that is, $r \le 33.$

Each odd row $r = 1, 3, \ldots, 33$ contains exactly one such entry, for a total of $17.$