12. (33rd American High School Mathematics Examination) A sequence of consecutive positive integers starting with 1 is written on the blackboard, and one of the numbers is erased. The arithmetic mean of the remaining numbers is $35 \frac{7}{17}$. The number that was erased is $\qquad$.
12. 7.

Brief solution: Let $n$ be the largest integer written on the blackboard.
If 1 is the number that is erased, then the maximum arithmetic mean is $\frac{2+3+\cdots+n}{n-1} = \frac{\frac{(n+1) n}{2}-1}{n-1} = \frac{n+2}{2}$;

If $n$ is the number that is erased, then the minimum arithmetic mean is $\frac{1+2+3+\cdots+n-1}{n-1} = \frac{n(n-1)}{2(n-1)} = \frac{n}{2}$.
Thus, $\frac{n}{2} \leqslant 35 \frac{7}{17} \leqslant \frac{n+2}{2}$, which means $n \leqslant 70 \frac{14}{17} \leqslant n+2$.
Therefore, $68 \frac{14}{17} \leqslant n \leqslant 70 \frac{14}{17}$. Hence, $n=69$ or 70.
Since $35 \frac{7}{17}$ is the arithmetic mean of $n-1$ integers, $\left(35 \frac{7}{17}\right)(n-1)$ must be an integer. When $n=70$, this is impossible, so $n=69$.

If $x$ is the number that is erased, then $\frac{\frac{69 \cdot 70}{2}-x}{68} = 35 \frac{7}{17}$, which means $69 \cdot 35 - x = 35 \frac{7}{17} \cdot 68 = 35 \cdot 68 + 28$, or $35 - x = 28$, solving for $x$ gives $x=7$.