Lattice point $(x, y)$ is visible from origin if $x$ and $y$ are relatively prime (or if $x^2+y^2=1$ but this case is not interesting). Of course number of good lines is visible points with integer coordinates $(x, y), 1 \leq x, y \leq n$ minus visible points with integer $(x, y), 1 \leq x, y \leq \frac{n}{2}$. Let $p$ be the probability that random point with integer coordinates $(x, y), 1 \leq x, y \leq n$ is visible from origin. We can see that probability that at least one coordinate is not divisible by $2$ is at least $1-\frac{1}{2^2}$. Analogously for other primes which are at most $n$. So $p \geq (1-\frac{1}{2^2})(1-\frac{1}{3^2})(1-\frac{1}{5^2})...(1-\frac{1}{p_k^2})$, where $p_k$ is largest prime number such that $p_k \leq n$. But $(1-\frac{1}{2^2})(1-\frac{1}{3^2})(1-\frac{1}{5^2})...(1-\frac{1}{p_k^2}) \geq (1-\frac{1}{2^2})(1-\frac{1}{3^2})(1-\frac{1}{4^2})...(1-\frac{1}{n^2})=\frac{1 \cdot 3}{2 \cdot 2} \frac{2 \cdot 4}{3 \cdot 3} ... \frac{(n-1)(n+1)}{n \cdot n}=\frac{n+1}{2n} >\frac{1}{2}$ This gives us that there are more than $\frac{n^2}{2}$ visible points with integer coordinates $(x, y), 1 \leq x, y \leq n$ and visible points with integer $(x, y), 1 \leq x, y \leq \frac{n}{2}$ is at most $\frac{n^2}{4}$ so number of good lines is more than $\frac{n^2}{2}-\frac{n^2}{4}=\frac{n^2}{4}$. If I'm not mistaken it's possible to prove stronger inequality which gives us that number of good lines is at least $\frac{n^2}{3}$.

Of course this solution isn't correct because events aren't independent :<. But it must be very close...