Hello, $\left ( 10^{1337} + \frac{1}{3} 10^{134} - \frac{1}{3} \right ) ^2$ does the job quite easily.

Denote ${x_k} = \frac{{{{10}^k} + 5}}{3} = \underbrace {33...3}_{k - 1}5$, then $x_k^2 = {\left( {\frac{{{{10}^k} + 5}}{3}} \right)^2} = \underbrace {11...1}_{k - 1}\underbrace {22...2}_k5$, $S\left( {x_k^2} \right) = 3k + 4$, So $S\left( {x_{669}^2} \right) = 3 \cdot 669 + 4 = 2011$. For any $n = S\left( {{x^2}} \right)$, we have $n \equiv {x^2}(\bmod 9)$, so $n \equiv 0(\bmod 9)$ or $n \equiv 1(\bmod 3)$. If $n = 9k$, denote ${y_k} = \frac{{{{10}^k} - 1}}{3} = \underbrace {33...3}_k$, $y_k^2 = \underbrace {11...1}_{k - 1}0\underbrace {88...8}_{k - 1}9$, $S\left( {y_k^2} \right) = 9k = n$. If $n = 3k + 1$, for $k = 0$, $S\left( {{1^2}} \right) = 3k + 1 = n$; for $k = 1$, $S\left( {{2^2}} \right) = 3k + 1 = n$; for $k \ge 2$, denote ${x_k} = \frac{{{{10}^k} + 5}}{3} = \underbrace {33...3}_{k - 1}5$, $x_k^2 = {\left( {\frac{{{{10}^k} + 5}}{3}} \right)^2} = \underbrace {11...1}_{k - 1}\underbrace {22...2}_k5$, $S\left( {x_{k - 1}^2} \right) = 3k + 1=n$.

$99999...999997$, where we have $222$ "nines" satsifies the condition.