This is very easy! (Not really)

$\color{#3D99F6} { \begin{array}{ccccc}\\ &Th&h&t&u\\ \hline &T&H&I&S\\ &V&E&R&Y\\ \hline &E&A&S&Y\\ \hline \end{array} }$
$u:~~\\ S+Y=Y,~~\implies~~S=0,~~and~NO~carry~to~t.\\ ~~~~~\\ t:~~\\ \therefore~~I+R=\overline{1S}=\overline{10},\\ \implies~~Also~there~is~a~carry~of~1~from~t~to~h.\\ ~~~~~\\ Th.~and~h:~~ \\ To~maximize,~ E=9,~or~8,~or~7,~. . .priority~in~descending~order.\\ With~E=9 ~~of~Th,~E=9~~of~h~also.\\ But~then~H=A,~since~there~is~a~carry~from~t.~~~\therefore~E=9.~Invalid.\\ Next~E=8.\\ To~maximize,~ A=9,~or~7,~or~6,~. . .priority~in~descending~order.\\ ~~~~~\\ h:~\\ There~is~a~carry~in~and~carry~out.\\ If~\{E,~ A\}=\{8,~9\},~but~then~since~there~is~a~carry~in,~~\therefore~H=0.~~\therefore~A=9~Invalid.\\ Next~\{E,~ A\}=\{8,~7\},~~but~H~also~=8=E.~\therefore~A=7~~Invalid. \\ Next~\{E,~ A\}=\{8,~6\},~and~H=7.~~\{A,~E,~H\}=\{6,8,7\}~valid.\\ ~~~~~\\ Back~to~Th:~\\ E=8,~and~there~is~a~carry.\\ \therefore~T+V=7.~A=6.~~So~T+V =3+4~~or~~=2+5,~in~any~order. \\ ~~~~~\\ Back~to~t:~\\ We~are~left~only~with~I+R=1+9,~in~any~order.\\ ~~~~~\\ Back~to~u:~\\ Option~left~for~Y~is,~~2,~or~3,~or~4,~or~5.\\ To~maximize,~Y=5,~~then~T+V=3+4,~in~any~order.\\ ~~~~~\\ \implies~~\overline{EASY}=\Large~~\overline{\color{#D61F06}{8605}}.\\$

The first thing that is easy to see here is,S=0, as anything added to itself is itself.So this simply implies no carry to the count of ten's digit.We can also see S in the answer as a tens digit, which then implies the compulsory carry of 1(only 1 is possible) to the count of hundred's digit.Now if we see the 3rd column H, E,A, we can easily conclude that none of them is 9, otherwise either H and A or E and A would be same digit.Now in order to maximize the sum, the best shot is to choose E=8 and H=7.This left us with one pair that sums to ten i.e. either of R or I is 1 or 9.A=6 is the result of supposing H and E, with carry over to thousand's digit count.Now here we have a choice of 4 or 3 as T or V in order to get the sum to 8.Now for the value of Y we have choices two choices, 2 or 5, and for the given maximizing problem, 5 fits the criterion.And so the final EASY=8605.

It was actually pretty easy with code:

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from itertools import permutations
for t,h,i,s,v,e,r,y,a in permutations('0123456789',9):
  if '0' in (t,v,e):
    continue
  if int(t+h+i+s)+int(v+e+r+y)==int(e+a+s+y):
    print(e+a+s+y)