1. In the Cartesian coordinate system, a point whose both horizontal and vertical coordinates are integers is called an integer point, such as $(-1,7)$ is an integer point. If the line $l$ passes through the points $A\left(\frac{1}{2}, \frac{1}{3}\right)$ and $B\left(\frac{1}{4}, \frac{1}{5}\right)$, then the integer point on $l$ that is closest to point $A$ is
Ni.1. $(-2,-1)$.
Suppose the equation of line $l$ is $y=k x+b$. According to the problem, we have
$$
\left\{\begin{array}{l}
\frac{1}{3}=\frac{1}{2} k+b, \\
\frac{1}{5}=\frac{1}{4} k+b .
\end{array}\right.
$$

Solving these equations, we get $k=\frac{8}{15}, b=\frac{1}{15}$.
Thus, the equation of line $l$ is $y=\frac{8}{15} x+\frac{1}{15}$.
Clearly, $(-2,-1)$ lies on $l$.
Substituting $x=-1,0,1,2,3$ into $y=\frac{8}{15} x+\frac{1}{15}$, we find that the corresponding $y$ values are not integers. Therefore, the nearest integer point to point $A$ on $l$ is $(-2,-1)$.