17. On the blackboard, there are $n$ consecutive positive integers starting from 1. After erasing one of these numbers, the average of the remaining numbers is $36 \frac{2}{5}$. The number that was erased is $\qquad$ .
17. 8. If the smallest number 1 is erased, then the average of the remaining numbers is $\frac{2+3+\cdots+n}{n-1}=\frac{n+2}{2}$. If the largest number is erased, then the average of the remaining numbers is $\frac{1+2+3+\cdots+(n-1)}{n-1}=\frac{n}{2}$, thus $\frac{n}{2} \leqslant 36 \frac{2}{5} \leqslant \frac{n+2}{2}$, which gives $70 \frac{4}{5} \leqslant n \leqslant 72 \frac{4}{5}$. Since $n$ is an integer, only $70 \leqslant n \leqslant 72$. And since the average of $n-1$ integers is $36 \frac{2}{5}$, $n-1$ must be a multiple of 5, which can only be $n=71$. Let the erased number be $x$, then $\frac{1+2+\cdots+71-x}{71-1}=36 \frac{2}{5}$, solving for $x$ gives $x=8$.