• Solve $(b)$ first.
• For part $(b)$ one is divisible by 100. In the next 79 integers, one has sum divisible by 17. In the previous 80 integers, one has sum divisible by 17.

$a)$ The sequence is $[999999999999921,~1000000000000078]$

Sum of digits of $999999999999921$ is 1 modulo 17 and sum of digits of $999999999999999$ is 16 modulo 17. Every integer in between has sum between 1 and 16 modulo 17. So, none of these sums is divisible by 17.

Sum of digits of $1000000000000000$ is 1 modulo 17 and sum of digits of $1000000000000078$ is 16 modulo 17. Every integer in between has sum between 1 and 16 modulo 17. So, none of these sums is divisible by 17.

$b)$ In any 100 consecutive integer, one of them is divisible by 100.

So, there is an integer divisible by 100 in the series.

Let the integer be $k=\overline{...00}=\overline{X00}$.

If there are at least 79 more integers after $k$ in the set, the integers:

$\overline{X00},~\overline{X01},~\overline{X02},~\overline{X03},~\overline{X04},~\overline{X05},~\overline{X06},~\overline{X07},~\overline{X08},~\overline{X09},~\\\overline{X19},~\overline{X29},~\overline{X39},~\overline{X49},~\overline{X59},~\overline{X69},~\overline{X79}$

have different sum of digit modulo 17.

But, there are 17 integers here. So, one of them has sum of digits divisible by 17.

So, there is not 79 integers after $k$ in the set.

So, there is at least $159-1-78=80$ integers in the set before $k$.

Let the number immediately before $k$ be $\overline{...99}=\overline{Y99}$.

The integers:

$\overline{Y99},~\overline{Y98},~\overline{Y97},~\overline{Y96},~\overline{Y95},~\overline{Y94},~\overline{Y93},~\overline{Y92},~\overline{Y91},~\overline{Y90},~\\\overline{Y80},~\overline{Y70},~\overline{Y60},~\overline{Y50},~\overline{Y40},~\overline{Y30},~\overline{Y20}$

have different sum of digit modulo 17.

But, there are 17 integers here. So, one of them has sum of digits divisible by 17. Proved.