Answer: 248

Small Hint:

Since $m$ and $n$ are odd, the midpoint of the centers of squares $200$ and $2000$ is exactly the center of square $1100,$ one square to the right of square $1099$

Big Hint:

The line enters square $1099$ precisely when its slope is less than $1$ in absolute value; then count the odd $n$ for each $m \in \{3, 5, 7, 9\}$

Solution:

Use column-row coordinates. Square $200$ is in the top row, so $n \ge 200$ (hence $n \ge 201$ as $n$ is odd) and its center is $(200, 1).$ Square $2000$ is in the bottom row, so its column is $b = 2000 - (m-1)n$ with $1 \le b \le n,$ i.e. $(m-1)n \lt 2000 \le mn,$ and its center is $(b, m).$ Since $m$ and $n$ are odd, $b$ is even, so the midpoint $\left(\frac{200 + b}{2}, \frac{m+1}{2}\right)$ of the two centers has integer coordinates; its square number is $\frac{(m-1)n + 200 + b}{2} = \frac{2200}{2} = 1100.$ Its column $100 + \frac{b}{2}$ lies between $101$ and $n,$ so the line passes through the center of square $1100,$ and square $1099$ sits immediately to its left in the same row.

Square $1099$ lies only in that row, and the line crosses that row’s horizontal strip in a segment centered (by symmetry) at the center of square $1100,$ extending $\frac{1}{2|s|}$ to each side, where $s = \frac{m-1}{b - 200}$ is the slope. So the line meets the interior of square $1099$ exactly when $\frac{1}{2|s|} \gt \frac{1}{2},$ that is $|s| \lt 1,$ i.e. $m - 1 \lt |b - 200|$ $= |1800 - (m-1)n|$ (a vertical line, $b = 200,$ fails).

Since $n \ge 201$ and $(m-1)n \lt 2000,$ we need $m - 1 \lt 10,$ so $m \in \{3, 5, 7, 9\}.$ For $m = 3:$ odd $n \in [667, 999],$ $167$ values, excluding $|n - 900| \le 1$ (the odd cases $899$ and $901$) leaves $165.$ For $m = 5:$ odd $n \in [401, 499],$ $50$ values, excluding $449$ and $451$ leaves $48.$ For $m = 7:$ odd $n \in [287, 333],$ $24$ values, excluding $299$ and $301$ leaves $22.$ For $m = 9:$ odd $n \in [223, 249],$ $14$ values, excluding $225$ leaves $13.$ The total is $165 + 48 + 22 + 13 = 248.$