Solution:
The main idea is to visit odd numbers!
We claim inductively that the first $2^{n}$ rows of Pascal's triangle satisfy the following properties:
- The $2^{n}$th row contains only odd numbers.
- The first $2^{n}$ rows contain $3^{n}$ odd numbers.
- When taken modulo 2, there is 120 degree symmetry
- There is a path starting at any corner to any other corner through only odd numbers.
Indeed this is clear for $n=1$. For the inductive step, let $T$ denote the shape of the first $2^{n}$ rows modulo 2. Note that row $2^{n}+1$ contains all even numbers except the endpoints $\left(\begin{array}{c}2^{n} \\ 0\end{array}\right)=\left(\begin{array}{c}2^{n} \\ 2^{n}\end{array}\right)=1$. Thus in fact we get two side-by-side copies of the triangle $T$, which meet on row $2^{n+1}$ and thus have all ones. (Between the two copies of $T$ we get an inverted triangle having all entries 0.) From these observations, we see that all statements in the inductive hypothesis hold.
Thus, we may visit odd numbers from rows 1 to 128. In doing so, we visit $3^{7}=2187$ odd numbers.