Note that only 4 3-digit "winning" numbers have squares that end in 444:

038^2 = 1,444

538^2 = 289,444

462^2 = 213,444

962^2 = 925,444

So, a low effort brute force attack is to take 444...444 starting with 0 intervening digits, then one intervening digit, then two etc, both for all 0's and for all 9's; take the square roots, and by inspection see if there is a number between those 2 square roots that matches one of our 4 three-digit winners above.

Example:

sqrt of 444000444 is 21,071.3180 etc

sqrt of 444999444 is 21,095.0099 etc

of the 23 numbers between 21,071 and 21,095, none are "winners"

But when there are 6 intervening numbers (444000000444 and 444999999444) we get two square roots that are

666,333.25 etc and 667,083.20 etc

The smallest three-digit winner (which is larger than 333) is 462.

So our square root is 666,462.

Squaring this gives 444,171,597,444