Following the hint, we observe that the ball lands in a pocket as soon as the line segment reaches a point $(c, c)$ where $c$ is divisible by both $a$ and $b$; clearly $c$ is the least common multiple of $a$ and $b$. The ball touches a vertical rail $c$ times and a horizontal rail $c$ times. If $c$ is even/odd, the ball lands in a pocket on the left/right side of the table, and if $c$ is even/odd, the ball lands in a pocket on the lower/upper side of the table. Hence the ball lands in the upper left pocket exactly when $c$ is even and $c$ is odd.

Write $a = 2^j a'$ and $b = 2^k b'$, where $a'$ and $b'$ are odd. Then $c = 2^{\max(j,k)} \operatorname{lcm}(a', b')$. Since $\operatorname{lcm}(a', b')$ is odd, $c$ is even and $c$ is odd precisely when $j < k$.

If $a$ is odd, then the ball lands in the upper left pocket whenever $b$ is even; there are $\frac{a-1}{2}$ such allowed values of $b$. If $a$ is divisible by $2$ but not by $4$, then the ball lands in the upper left pocket whenever $b$ is divisible by $4$; there are $\frac{a-2}{4}$ such allowed values of $b$, and so forth.

Fix an integer $N$ and count the ordered pairs $(a, b)$ with $1 \le b \le a \le N$ for which the ball lands in the upper left pocket. In the calculation we use $C(k, 2) = \binom{k}{2} = 0 + 1 + 2 + \dots + (k-2) + (k-1)$.

When $a = 1, 3, 5, 7, \dots$, then there are $0, 1, 2, 3, \dots$ allowed values of $b$; summing over all such $a$ gives us $C\left(\left[\frac{N+1}{2}\right], 2\right)$ total ordered pairs. When $a = 2, 6, 10, 14, \dots$, there are $0, 1, 2, 3, \dots$ allowed values of $b$; summing over all such $a$ gives us $C\left(\left[\frac{N+2}{4}\right], 2\right)$ total ordered pairs, and so forth. For $N = 2009$, there are a total of $C(1005, 2) + C(502, 2) + C(251, 2) + C(126, 2) + C(63, 2) + C(31, 2) + C(16, 2) + C(8, 2) + C(4, 2) + C(2, 2) = 672,084$ allowed ordered pairs, so this is our answer.