Solution:

Shade three of the six rows in column $1:$ $\binom{6}{3} = 20$ ways. Let $k$ be the number of rows shaded in both columns $1$ and $2;$ column $2$ can then be chosen in $\binom{3}{k}\binom{3}{3-k}$ ways. After these two columns, $k$ rows are complete with two shaded squares, $6 - 2k$ rows have one, and $k$ rows have none.

The empty rows must be shaded in both columns $3$ and $4.$ Column $3$ takes those $k$ rows plus $3 - k$ of the $6 - 2k$ singly-shaded rows, in $\binom{6-2k}{3-k}$ ways, and column $4$ is then forced: it must cover the empty rows and exactly the singly-shaded rows skipped by column $3.$

Summing, $$N = 20\sum_{k=0}^{3} \binom{3}{k}\binom{3}{3-k}\binom{6-2k}{3-k} = 20(20 + 54 + 18 + 1) = 1860,$$ so the remainder is $860.$