Answer:$36$ Example (Sketch): Construct a $9x9$ table where the squares in this table are the vertices in the $8x8$ table. There is no diagonals which pass through $1,3,5,7,9.$ squares on the last row and last column. Proof: Enumerate the vertices as $1,2,3,4$. $\begin{bmatrix} 1 2 1 2 1 2 1 2 1\\ 343434343 \\ 1 2 1 2 1 2 1 2 1\\ 343434343 \\ 1 2 1 2 1 2 1 2 1\\ 343434343 \\ 1 2 1 2 1 2 1 2 1\\ 343434343 \\ 1 2 1 2 1 2 1 2 1\\ \end{bmatrix}$ $1$ and $4$ can be connected and $2$ and $3$ can be connected. The others are not possible. There are $25$ times $1$, $20$ times $2$, $20$ times $3$ and $16$ times $4$. So there can be at most $16$ diagonals between $1$ and $4$. There can be at most $20$ diagonals between $2$ and $3$. So in total, there may be at most $36$ diagonals as desired.$\blacksquare$