10. (20 points) The numbers $1, 2, \cdots, 2014$ are written on a blackboard. The following operation is performed: in the first step, the first two numbers $1$ and $2$ are erased, and their sum $3$ is written at the end of the sequence; in the second step, the first three numbers $3$, $4$, and $5$ are erased, and their sum $12$ is written at the end of the sequence; and so on. The operation continues until, in the $t$-th step, there are fewer than $t+1$ numbers left on the blackboard, at which point the operation stops. Find the number of distinct numbers that have appeared on the blackboard and the sum of these distinct numbers (if a number appears multiple times, it is counted only once).
10. Since after the $k$-th step, the number on the blackboard decreases by $k$, hence
$$
\begin{array}{l}
1+2+\cdots+(t-1)2014 .
\end{array}
$$

Therefore, the numbers written in the first 14 steps are all less than 2014, and the numbers written afterward are all greater than 2014.

The numbers erased in the 62nd step are $1953, 1954, \cdots, 2014$ (62 numbers) and $3(=1+2)$, and the number written is
$$
62 \times \frac{1953+2014}{2}+3=122980 \text {. }
$$

The new numbers written are all distinct, so the total number of different numbers on the blackboard is
$$
2014+62-14=2062 \text { (numbers). }
$$

The initial numbers on the blackboard are $1,2, \cdots, 2014$, and the sum of the remaining numbers is $1+2+\cdots+2014$. Among them, $\frac{k(k+1)(k+2)}{2}(k=2,3, \cdots, 14)$ have already appeared, so the sum of the different numbers that have appeared on the blackboard is
$$
\begin{array}{l}
2014 \times 2015-\sum_{k=2}^{14} \frac{k(k+1)(k+2)}{2} \\
=2014 \times 2015-\sum_{k=2}^{14}\left\{\sum_{i=0}^{k}\left[\frac{k(k+1)}{2}+i\right]\right\} \\
=2014 \times 2015-\sum_{i=3}^{119} i \\
=4058210-117 \times 61=4051073 .
\end{array}
$$