We will prove that there is no tri-connected collection if $n$ is odd, and that tri-connected collections exist for all even $n \geq 38$. Since there are 501 even numbers in the range from 2018 to 3018, this yields 501 as the answer. For any two different squares $A$ and $B$, let us write $A \sim B$ to mean that square $A$ touches square $B$. Since each square touches exactly three other squares, and there are $n$ squares in total, the total number of instances of $A \sim B$ is $3 n$. But $A \sim B$ if and only if $B \sim A$. Hence the total number of instances of $A \sim B$ is even. Thus $3 n$ and hence also $n$ is even. We now construct tri-connected collections for each even $n$ in the range. We show two constructions. Construction 1: The idea is to use the following two configurations. Observe that in each configuration every square is related to three squares except for the leftmost and rightmost squares which are related to two squares. Note that the configuration on the left is of variable length. Also observe that multiple copies of the configuration on the right can be chained together to end around corners. Putting the above two types of configurations together as in the following figure yields a tri-connected collection for every even $n \geq 38$. Construction 2: Consider a regular $4 n$-gon $A_{1} A_{2} \cdots A_{4 n}$, and make $4 n$ squares on the outside of the $4 n$-gon with one side being on the $4 n$-gon. Reflect squares sharing sides $A_{4 m+2} A_{4 m+3}, A_{4 m+3} A_{4 m+4}$ across line $A_{4 m+2} A_{4 m+4}$, for $0 \leq m \leq n-1$. This will produce a tri-connected set of $6 n$ squares, as long as the squares inside the $4 n$-gon do not intersect. When $n \geq 4$, this will be true. To treat the other cases, consider the following gadget: Two squares touch 3 other squares, and the squares containing $X, Y$ touch 2 other squares. Take the $4 n$-gon from above, and break it into two along the line $A_{1} A_{2 n}$, moving the two parts away from that line. Do so until the gaps can be exactly filled by inserting two copies of the above figure, so that the vertices $X, Y$ touch the two vertices which used to be $A_{1}$ in one instance, and the two vertices which used to be $A_{2 n}$ in the other. This gives us a valid configuration for $6 n+8$ squares, $n \geq 4$. Finally, if we had instead spread the two parts out more and inserted two copies of the above figure into each gap, we would get $6 n+16$ for $n \geq 4$, which finishes the proof for all even numbers at least 36.