Solution sketch: The construction is an isometric (triangular) grid, bounded by a hexagon with $n$ points on each side. It is easily calculated that there are $3n^2-3n+1$ points. Let us now compute the distances. Denote the hexagon by $ABCDEF$. For every segment in the hexagon, we may translate it so that one of the endpoints is $A$. Then, we can rotate the segment a multiple of $60^{\circ}$ so that its other endpoint lies in the angle $\angle BAD$. Finally, we can reflect the segment across $AC$ if necessary to make the other endpoint lie in $\bigtriangleup ACD$. It is easy to see that this always works. [asy][asy] pair A = dir(180); pair B = dir(120); pair C = dir(60); pair D = dir(0); pair E = dir(300); pair F = dir(240); draw(A--B--C--D--E--F--A--cycle); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); draw(C--A--D); [/asy][/asy] There are a total of $n^2$ lattice points in triangle $ACD$. Hence, the $3n^2-3n+1$ lattice points in the hexagon define at most $n^2$ distances. Clearly $3n^2-3n+1 > 2n^2+2020$ for large $n$.