jmc

Let the desired number be $a$. Then $$\begin{array}{rl}a\equiv 0& (\mathrm{mod}17)\\ a\equiv -1\equiv 7& (\mathrm{mod}8)\end{array}$$ The first congruence implies that there exists a non-negative integer $n$ such that $a=17n$. Substituting this into the second congruence yields $$17n\equiv 7(\mathrm{mod}8),$$ $$⟹n\equiv 7(\mathrm{mod}8).$$ So $n$ has a lower bound of $7$. Then $$n\ge 7,$$ $$⟹a=17n\ge 119.$$ $119$ satisfies both congruences, so subtracting it from both sides of both congruences gives $$\begin{array}{rl}a-119\equiv -119\equiv 0& (\mathrm{mod}17)\\ a-119\equiv -112\equiv 0& (\mathrm{mod}8)\end{array}$$ Since $\gcd (17,8)=1$, we get $a-119\equiv 0(\mathrm{mod}17\cdot 8).$ That is, $a\equiv 119(\mathrm{mod}136).$

Note that every number that satisfies this congruence satisfies the original two congruences. The largest number of the form $119+136m$ for some non-negative integer $m$, and less than $1000$, is $119+136\cdot 6=\boxed{935}.$