Let $m$ be the least positive integer divisible by $17$ whose digits sum to $17$. Find $m$. 
Let $m$ be the least positive integer divisible by $17$ whose digits sum to $17$. Find $m$.

$m$ is a 3 digit number (because this was an AIME problem). 
$$m \equiv 0 \pmod{17}$$
$$m \equiv 17 \pmod{9} \equiv -1 \pmod{9}$$
Applying the chinese remainder theorem, the solution is supposed to be:
$$m = 17 \cdot 9 = 153$$
but that isnt correct. 

Lebes, from your lead and solving for $N$, we have
$$N \equiv 0 \pmod{17}\equiv 17 \pmod{17}$$
and from the digit sum condition
$$N \equiv 17 \pmod{9}.$$
Since $(17,9)=1$, we have $N\equiv17\pmod{153}$. [This uses principles from the Chinese Remainder Theorem - see the bottom for a detailed explanation]
Since this is only finding solutions where the digitsum $\equiv8\pmod{9}$, we now simply check the first few solutions to find the correct answer:
$1\times 153+17=170\longrightarrow $ digit sum  $=8\qquad$Nope.
$2\times 153+17=323\longrightarrow $ digit sum  $=8\qquad$Nope. [Fixed the error here]
$3\times 153+17=476\longrightarrow $ digit sum  $=17\qquad$Found it!
Therefore $N=476$.
Thanks @AaronMaroja for the correction.

Explanation of the use of CRT for @Amad27:
Using the Chinese Remainder Theorem to solve $N \equiv 0 \pmod{17}$ and $N \equiv 17 \pmod{9} \equiv 8 \pmod{9}$. Breaking these down we get $a_1=0$, $a_2=8$, $m_1=17$, $ m_2=9$, $M=m_1\times m_2=153$, $M_1=M/m_1=153/17=9$, $M_2=M/m_2=153/9=17$. 
Now applying CRT we get
$$M_1y_1\equiv 1\pmod{m_1}\rightarrow 9y_1\equiv 1\pmod{17},$$
and
$$M_2y_2\equiv 1\pmod{m_2}\rightarrow 17y_2\equiv 1\pmod{9}\rightarrow 8y_2\equiv 1\pmod{9}.$$
It can be seen by observation that $y_1=2$ and $y_2=8$.
Finally, combining the above information we get
$$m=a_1M_1y_1+a_2M_2y_2=0\times9\times2+8\times17\times8=1088.$$
Since $1088\equiv17\pmod{153}$, we have
$$N\equiv m\pmod{M}\equiv17\pmod{153}$$
