Gridlock

To establish an upper bound, note that we could use the first 25 positive even integers in any configuration, in which case the gcd of adjacent numbers will always exceed 1. This gives us an upper bound of $50$ for the smallest maximum. Also, we cannot include $1$ , as the gcd of it and any other positive integer will be 1. Thus at best we would be using the integers $2$ through $26$ .

However, consider what happens with primes $\ge 17$ . Each square is adjacent to at least two other squares, so any number included must have two other numbers with which it shares a non-unit gcd. With $17$ , we would have to include $34$ and $51$ at best, but as we know that $50$ is an upper bound we know that $17$ cannot be part of an optimal solution. The same goes for $19$ and $23$ . So at this point the best we could do is find a solution that involves the first 25 positive integers greater than 1 and excluding $17, 19$ and $23$ . This would raise our largest integer from $26$ to $29$ , but then $29$ would have the same issue as $17$ , so we would drop $29$ and add $30$ .

But now we have to deal with $11$ and $13$ . If we include $11$ , we would have to include $33$ , (with $22$ already in play), and if we include $13$ we would have to include $39$ , (with $26$ already in play). If we dropped $11$ and $13$ then the next available integers to add to the list would be $32$ and $33$ , (since $31$ would have the same issues as $17$ ). This is one possibility, but if we keep $11$ and only drop $13$ then we would have to add $33$ as the 25th integer for reasons already discussed. So either way the best we can do is to work with a group of 25 integers the maximum of which is $33$ , with the latitude to drop one otherwise suitable number less than $33$ from the grid. So if we can create a suitable grid with this group of 25 integers, (with one degree of flexibility), then the smallest maximum would be $\boxed{33}$ . Here is such a grid, (with $27$ chosen to be excuded):

$\Large 7 \space \space 21 \space \space 18 \space \space 15 \space \space 3$

$\Large 14 \space 28 \space \space 2 \space \space 12 \space \space 9$

$\Large 4 \space \space \space 8 \space \space 16 \space \space 26 \space \space 24$

$\Large 10 \space 20 \space \space 32 \space \space 6 \space \space 22$

$\Large 5 \space \space 25 \space \space 30 \space \space 33 \space \space 11$