Sum of digits

As $n$ is a positive integer so $n^2=\overline{a_k a_{k-1} a_{k-2} .....a_2 a_1}$ .

$n^2=10^{k-1} a_k +10^{k-2} a_{k-1} +......+ 10 a_2 +a_1$

$\implies n^2=9 l + \sum_{r=1}^{k} a_r$

$\implies n^2=9 l +2000$

So it seems there is no solution but if $9 l$ ends with $8000$ there may exists a solution.

Consider $9 l=\overline{X8000}$ where $X$ is any integer So, $1000\, |\, l$ where

$l=(\underbrace{111....11}a_k + \underbrace{111...11} a_{k-1} + .....+111 a_4+11a_3 +a_2$

$l=\underbrace{111...000}a_k + \underbrace{111...000}a_{k-1} + ....+ 111a_k +111a_{k-1}+...111a_4 +11a_3+a_2$

$l=1000m +111(a_k +a_{k-1}+...+a_5+a_4)+11 a_3 +a_2$

$l=1000m +111(2000 -a_3-a_2-a_1)+11 a_3+a_2$

$l=1000m' -100 a_3 -110a_2-111 a_1 =1000m' -100(a_3+a_2+a_1)-10a_2-11 a_1$

Now $1000 | 100(a_3+a_2+a_1)+10a_2+11 a_1$ . The first two term ends with zero but last does not.if $a_1=0$ then $a_2=0$ beacuse $n^2$ is perfect square..So $a_3$ should be $10$ which is not possible.So , $1000 \not | l$ .

So our assumption is wrong no such $n$ exists.