Let $N$ be the largest positive integer with the following property: reading from left to right, each pair of consecutive digits of $N$ forms a perfect square. What are the leftmost three digits of $N$?
The two-digit perfect squares are $16, 25, 36, 49, 64, 81$. We try making a sequence starting with each one:

$16 - 64 - 49$. This terminates since none of them end in a $9$, giving us $1649$.
$25$.
$36 - 64 - 49$, $3649$.
$49$.
$64 - 49$, $649$.
$81 - 16 - 64 - 49$, $81649$.
The largest is $81649$, so our answer is $\boxed{816}$.