Case2
$k$ is a odd number
$\therefore 4k^{2}\equiv 1(mod17)$
$(2k+1)(2k-1)=17m$
2.1$2k+1=17s$
$\therefore n=17s-1$
all n in this case is $\left\lceil\ \frac{1001}{17} \right\rceil=58$
2.2$2k-1=17t$
$\therefore n=17t+1$
all n in this case is $\left\lceil\ \frac{999}{17} \right\rceil=58$
from Case1,2 $\therefore$ all n is $58+58+58=174$
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