Note that the triangle with vertices $A(0, 0), B(63, 0),$ and $C(0, 63)$ has exactly $64$ lattice points on each one of its sides whilst it's area is $\frac{63^2}{2} < 2020$. Assume there exists a lattice triangle with $m \ge 65$ lattice points on each one of its sides (with area less than 2020). WLOG assume we have $y_A < y_B$ and $x_A < x_B$, with the lattice points on $AB$ being $M_1, ..., M_m$, from where we get that, since $M_1$ is the closest lattice point to $A$, then the next lattice point, $M_2$, is $(x_A + (x_{M_1} - x_A), y_A + (y_{M_1} - y_A))$, from where we get that $AM_1 = M_1M_2 = ... = M_mB$. Similarly we can obtain $AN_1 = ... = N_mC$ (where $N_1, ..., N_m$ are the lattice points on $AC$), from where we get that $P_{AM_1C} = ... = P_{M_mBC}$ and $P_{AM_1N_1} = ... = P_{N_mM_1C}$, so $P_{ABC} = (m - 1)^2 \cdot P_{AM_1N_1}$. However since $AM_1N_1$ is a lattice triangle, using Pick's theorem we obtain $P_{AM_1N_1} = \frac{b}{2} + i - 1 \ge \frac{b}{2} - 1 = \frac{3}{2} - 1 = \frac{1}{2}$, so $P_{ABC} = (m - 1)^2 \cdot P_{AM_1N_1} \ge 64^2 \cdot \frac{1}{2} > 2020$, yielding a contradiction. Therefore, the maximum value of $m$ is $64$.