First, $N+N+Y$ yields $Y$.
So $N+N \in \{0,10\}$ which means that $N \in \{0,5\}$.
If $c_1$ is the carry over from the ones column into the tens, then $c+E+E+T \implies T$.
This means that $c_1+E+E \in \{0,10\}$. This is impossible unless $c_1=0$. Therefore, $N=0$ to get no carry over, and $E=5$ since 0 is taken. The carry over into the hundreds column is then 1.
We also know that
$1+T+T+R$ can carry over at most 2 if $T$ and $R$ are maxed. Lets call this carry over $c_2$. Thus, $O+c_2$ carries over, and it can be at most 1. Thus, $O\in \{8,9\}$, $I\in \{0,1\}$, and $S=F+1$.
But since
0 is already taken, we know $I=1$. Thus, $O=9$ and $c_2=2$. The lowest value for $X$ is then 2. Thus, $1+T+T+R \ge 22$. If $T=8$, then $R \in \{6,7\}$. If $T=7$, then $R=8$. $T=6$ does would require $R=9$ which is already taken.
We know that $S$ and $F$ must be
consecutive, and 7 is already taken. thus, $S \in \{3,4\}$ and $F \in \{2,3\}$. In both cases, 3 is taken, so $T=8, R=6$ cannot work because that would make $X=3$. Similarly, $T=7,R=8$ makes $X=3$, so that is also not an option.
Therefore,
$T=8$ and $R=7$ which makes $X=4$. Thus, $S=3$ and $F=2$. The only remaining value for $Y$ is 6.
Solution is:
0=N, 1=I, 2=F, 3=S, 4=X, 5=E, 6=Y, 7=R, 8=T, 9=O
850
+ 850
+ 39786
-------
41486