$\color{#D61F06}444$

We will search the interesting square numbers. The interesting numbers can be formed like $x^2=\overline{a_1a_2\dots a_n444}$ and let $y$ be $\overline{a_1a_2\dots a_n}$ . From that $x^2+1000y+444$ , so $x^2-1444=1000y-1000$ and $(x-38)(x+38)=1000(y-1)$ Since $1000|1000(y-1), 1000|(x-38)(x+38)$ . $1000=2^3*5^3$ .

If $(x-38)$ is not divisible by $4$ , then $(x+38)$ can't be divisible by $4$ too. The reverse of this is also true. So the product of them won't be a multiple of $1000$ . So $4|(x-38), (x+38)$ .

Now we will look at that the where are the $5$ s. There are two cases:

Case 1: $4|(x-38)$ and $4*125|(x+38)$ or the reverse.

Since $125*4=500$ , $x=500k\pm 38$ ( $k$ is an integer). $(500k\pm38)^2=250000k^2\pm38000k+1444,$ which ends in three $4$ s, so every $(500\pm38)^2$ is an interesting number.

Case 2: $4*5|(x-38)$ and $4*25|(x+38)$ or the reverse.

This means, that both numbers have to divisible by $10$ , but their difference is $76$ , which is impossible.

So the interesting numbers are formed like $(500k\pm38)^2$

Since $(500k-38)+(500k+38)=2*500$ and the first square root is $38$ , the answer is

$2*(1*500+2*500+\dots+\dfrac{101-1}{2}*500)=2*\dfrac{50*51}{2}*500=50*51*500=\boxed{1275000}$

So we want to find all $n^2\equiv444\ (\text{mod}\ 1000)$ .

This is simply $n^2\equiv4\ (\text{mod}\ 8)$ and $n^2\equiv69\ (\text{mod}\ 125)$ .

We find the square roots: $n\equiv2,6\ (\text{mod}\ 8)$ and $n\equiv38,87\ (\text{mod}\ 125)$ .

Thus, we have $n\equiv38,462,538,962\ (\text{mod}\ 1000)$ .

In other words, all interesting numbers are of the form $1000k+38, 1000k+462, 1000k+538, 1000k+962$ .

The answer is then basically then the sum of the first $100$ interesting numbers, add the $101$ st, and minus the $1$ st: $\sum_{k=0}^{24}(1000k+38)+\sum_{k=0}^{24}(1000k+462)+\sum_{k=0}^{24}(1000k+538)+\sum_{k=0}^{24}(1000k+962)+(1000\cdot25+38)-(38)$ $=\sum_{k=0}^{24}(4000k+2000)+1000\cdot25$ $=\frac{4000(24)(24+1)}{2}+2000\cdot25+1000\cdot25$ $=1275000$