Twice Year's Beginning

For $X^2$ to have starting digits of 20172017. This simply means that $20172017 \times 10^n \leq X^2 < 20172018 \times 10^n$

Now before taking square roots on both sides, we have to check that whether $n$ is odd or even. Considering both the cases:

$n \text{ odd: } 14202.8226068 \ldots \times 10^k \leq X < 14202.8229588 \ldots \times 10^k$ $n \text{ even: } 4491.32686408 \ldots \times 10^k \leq X < 4491.3269754 \ldots \times 10^k$

When $n$ is odd, as we vary $k$ , the smallest integer $X$ will be 142028227.

When $n$ is even, as we vary $k$ , the smallest integer $X$ will be $\boxed {44913269}$ .