This is true and I would be grateful if someone could make a figure to the below argument. Your conjecture follows from the following statement.

Let $\ell$ pass through the origin, $O$. For simplicity, suppose $\ell$ has a positive slope and let $P=(n,m)$ for some $m,n>0$ such that $P$ lies under $\ell$. Denote by $Q$ the point on $\ell$ whose $y$ coordinate is $m$ and by $R$ the point on the $x$-axis such that $ORPQ$ forms a parallelogram. Using symmetry, we get that if $OPQ$ is an empty triangle, then so if $ORP$.

Thus if $P$ minimizes the angle, there can be no closer points to the line in the lower part of the same quadrant. Notice also that all points in the quadrants that are not intersected by $\ell$ are farther than $P$.

This is true and I would be grateful if someone could make a figure to the argument below. Your conjecture follows from the following statement.

Let $\ell$ pass through the origin, $O$. For simplicity, suppose $\ell$ has a positive slope and let $P=(n,m)$ for some $m,n>0$ such that $P$ lies under $\ell$. Denote by $Q$ the point on $\ell$ whose $y$ coordinate is $m$ and by $R$ the point on the $x$-axis such that $ORPQ$ forms a parallelogram. Using symmetry, we get that if $OPQ$ is an empty triangle, then so if $ORP$.

Thus if $P$ minimizes the angle, there can be no closer points to the line in the lower part of the same quadrant. Notice also that all points in the quadrants that are not intersected by $\ell$ are farther than $P$.