Denote the following:
- $S_i$ to be the sum of the digits in column $i$.
- $D_i$ to be the digit in column $i$ of the sum. Thus $D_i = S_i \mod 10$
- $C_i$ to be the carry over of $S_i$.
Given that all the numbers are even, we know that $N,X,E \in \{0,2,4,6,8\}$. Looking at $S_1=N+N+X+E \le 8+8+6+4=26 \implies C_1 \le 2$.
Lets look at $S_2=U+U+U+Z+C_1$ and $D_2=Z$. We know that $3\times U + C_1 \mod 10 = 0$ in order to make this work. If $C_1=0 \implies U=0$, which can't happen. $C_1=1 \implies U=3$ and $C_1=2 \implies U=6$.
We can also see that $C_4 \le 1$ and since $S \ne D$, we have $C_4=1$ and $S=D+1$.
Now look at $S_1=N+N+X+E$ and $D_1=E$. This implies that $N+N+X \mod 10 = 0$. If $N=0$ then $X=0$, which can't happen.
- $N=2 \implies X=6$ and $C_1=1$ and $U=3$ and $C_2=1$
- $N=4 \implies X=2$ and $C_1=1$ and $U=3$ and $C_2=1$
- $N=6 \implies X=8$ and $C_1=2$ and $U=6$ which can't happen
- $N=8 \implies X=4$ and $C_1=2$ and $U=6$ and $C_2=2$
We also know that $C_3 \le 1$ and $C_4=1$ and $S=D+1$
Let assume $U=6$.
Thus, $N=8$, $X=4$, and $C_1=C_2=2$.
Then $S_3=E+6+2$. The valid choices for $E \in \{2,0\}$ If $E=0$ then $D_3=I=8$ which is already used. Thus $E=2$, and $I=0$ and $C_3=1$
In $S_4=D+O+C_3=D+O+1$ and $D_4=E=2$. Thus, $D+O=11$. But all the even numbers are already taken, so there is no way to make this equality work.
Thus $U \ne 6$.
Thus $U=3$
So we know that $N \in \{2,4\}$, $X \in \{6,2\}$, and $C_1=C_2=1$. In either case, $N$ or $X$ is $2$, so no other letter can be $2$. So $E \in \{0,4,6,8\}$.
If $E=8$, then $C_3=1$ and $S_4=D+O+C_3=D+O+1$ and $D_4=E$. This would require one of $D$ or $O$ to be 9, an the other 8.
If $E=4$, then $N=2$ and $X=6$. In $S_3=E+U+C_2=4+3+1=8$, so $D_3=I=8$ and $C_3=0$. From $S_4=D+O+C_3=D+O$ and $D_4=E=4$. The only combination that works are $D+O=9+5$. But either solution for $D$ results in an invalid $S$ since $S=D+1$.
If $E=0$, then from $S_3=E+U+C_2=0+3+1=4$ means that $D_3=I=4$ and $C_3=0$. Thus, $N=2$ and $X=6$. From $S_4=D+O+C_3=D+O$ and $D_4=E=0$. The only combinations of remaining numbers that work for $D$ and $O$ are $9$ and $1$. But neither result in a valid $S$ since $S=D+1$.
Thus $E=6$, then means that one of $D$ or $O$ is 8, and the other is 7. Since $S=D+1$, if $D=7$ then both $N=O=8$. Thus, $D=8$, $O=7$, and $S=9$. Also, $X=2$ and $N=4$. Also, from $S_3=E+U+C_2=6+3+1=10$, so $D_3=I=0$
The final solution is:
$$U=3, E=6, D=8, O=7, S=9, X=2, N=4, I=0, Z\in\{1,5\}$$