Solution:

Clearly $R = 1$, and from the hundreds column, $M = 0$ or $9$. Since $H + G = 9 + O$ or $10 + O$, it is easy to see that $O$ can be at most $7$, in which case $H$ and $G$ must be $8$ and $9$, so $M = 0$. But because of the tens column, we must have $S + T \geq 10$, and in fact since $D$ cannot be $0$ or $1$, $S + T \geq 12$, which is impossible given the remaining choices. Therefore, $O$ is at most $6$.

Suppose $O = 6$ and $M = 9$. Then we must have $H$ and $G$ be $7$ and $8$. With the remaining digits $0, 2, 3, 4$, and $5$, we must have in the ones column that $T$ and $S$ are $2$ and $3$, which leaves no possibility for $N$. If instead $M = 0$, then $H$ and $G$ are $7$ and $9$. Since again $S + T \geq 12$ and $N = T + 1$, the only possibility is $S = 8$, $T = 4$, and $N = 5$, giving $R O U N D = 16352 = 7004 + 9348 = 9004 + 7348$.