I take it that the points themselves do not have to have rational coordinates. I don't think that there would have to be a best approximation. Consider the two point set set $\{(0,0),(1,\frac{1}{\sqrt{2}})\}$. Then lines such as $y=\frac{12}{17}x$ and $y=\frac{408}{577}x$ based on convergents to $\frac{1}{\sqrt{2}}$ provide better and better approximations. Nothing changes if we use more points $(x_i,\frac{x_i}{\sqrt{2}})$ or use some other irrational slope.
Also, at the eight x values $\frac{\sqrt{3}-\sqrt{2}+j}{8}$ for $0 \le j \le 7$, the two functions $3x+\sqrt{2}$ and $-5x+\sqrt{3}$ are equal $\mod 1$ so either is a perfect fit.
later Based on some of the comments, here is one idea for a question: We are given some points $(x_i,y_i)$ in the unit square. For each integer $k \ge 0 $ consider all the lines $y=rx+b$ (with $r$ rational if desired, although it might be better to not make this restriction) such that $0 \le b <1$ and $k \le r <k+1$. For each such line find the squared distance to the points $(x_i,y_i+h_i$ where the $h_i$ are integers chosen to minimize each distance. QUESTION: Is there an elegant way to find the best approximating line(s) for each $k$ ? One should then also consider negative slopes.