Solution:
If Johann picks the point $(a, b)$, the path will contain $\gcd(a, 2004-b)+1$ points. There will be an odd number of points in the path if $\gcd(a, 2004-b)$ is even, which is true if and only if $a$ and $b$ are both even. Since there are $49^{2}$ points with $a, b$ both even and $98^{2}$ total points, the probability that the path contains an even number of points is

$$\frac{98^{2}-49^{2}}{98^{2}}=\frac{49^{2}\left(2^{2}-1^{2}\right)}{49^{2}\left(2^{2}\right)}=\frac{3}{4} .$$