6. Write a sequence of consecutive positive integers starting with 1 on the blackboard, and erase one of the numbers. The arithmetic mean of the remaining numbers is $35 \frac{7}{17}$. The number erased is
A. 7
B. 8
C. 9
D. Uncertain
6. Let $n$ be the largest integer written on the blackboard. If 1 is the number that is erased, then the largest arithmetic mean is $\frac{2+3+\cdots+n}{n-1}=\frac{(n+2)(n-1)}{2(n-1)}=\frac{n+2}{2}$; if $n$ is the number that is erased, then the smallest arithmetic mean is $\frac{1+2+\cdots+(n-1)}{n-1}=\frac{n(n-1)}{2(n-1)}=\frac{n}{2}$. Therefore, we have $\frac{n}{2} \leqslant 35 \frac{7}{17} \leqslant \frac{n+2}{2}$, solving this gives $68 \frac{14}{17} \leqslant n \leqslant 70 \frac{14}{17}$. Since $n$ is an integer, $n=69$ or 70.

Since $35 \frac{7}{17}$ is the arithmetic mean of $n-1$ integers, $35 \frac{7}{17}(n-1)$ must be an integer. When $n=70$, this is not satisfied, so $n=69$.
Let $x$ be the number that is erased, then $\frac{70 \cdot 69-2 x}{2 \cdot 68}=35 \frac{7}{17}$, solving this gives $x=7$ as the solution.