Going down:
For a positive integer $n$, let $s(n)$ denote the sum of its digits.
For any $x < 2021$, we will have $x^2 \leq 2020^2 < 10^7$ which means that $s(x^2) \leq 9 \times 7 = 63$.
Therefore, if we have $s(x^2) = s(x)^2$, then necessarily $s(x) \leq 7$.
Now just listing out all numbers $x$ with $s(x) \leq 7$, in descending order from $2020$, testing each of them until we get ten.
The answer is: 2020, 2012, 2011, 2010, 2002, 2001, 2000, 1301, 1300, 1220
Below are all the numbers being tested. There are $38$ numbers in total.
2020, 2014, 2013, 2012, 2011, 2010, 2005, 2004, 2003, 2002, 2001, 2000, 1600, 1510, 1501, 1500, 1420, 1411, 1410, 1402, 1401, 1400, 1330, 1321, 1320, 1312, 1311, 1310, 1303, 1302, 1301, 1300, 1240, 1231, 1230, 1222, 1221, 1220
Going up:
Same strategy, assuming that the results don't exceed $3000$ (which is verified a posteriori).
The answer is: 2022, 2100, 2101, 2102, 2110, 2111, 2120, 2121, 2200, 2201
A total of $25$ numbers are tested:
2022, 2023, 2030, 2031, 2032, 2040, 2041, 2050, 2100, 2101, 2102, 2103, 2104, 2110, 2111, 2112, 2113, 2120, 2121, 2122, 2130, 2131, 2140, 2200, 2201