Solution: This is a 3-digit number, so we can call the digits $ \text {A \ B \ C} $, where $ A \ne 0 $. This means: $ N = 100A + 10B + C $ and $ S = A + B + C $. So, by the definition of $M$, we have: \[ M = \dfrac {100A + 10B + C}{A + B + C}. \]Remark that this can be re-written as \[ M = \dfrac {99A + 9B}{A + B + C} + 1. \]Consider this as a function of $C$ and notice that $C$ only appears in the denominator. We can minimize this, so we want to maximize $C$; i.e. $C=9$. Substituting this into the equation above, we obtain: \[ M = \dfrac {99A + 9B}{A + B + 9} + 1 = \dfrac {(90A-81)+(9A+9B+81)}{A+B+9} + 1. \]Now, $M$ can be rewritten as \[ M = \dfrac {90A - 81}{A+B+9} + 9 + 1 = \dfrac {90A - 81}{A + B + 9} + 10. \]Again, we wish to minimize this so, in the same manner, we note that $B$ only appears in the denominator in this function of $B$. Clearly, we want to maximize $B$ to minimize this fraction. So, $B=9$. Substituting $B=9$ into this, we get: $ M = \dfrac {90A-81}{A+18} + 10 $. Note that \[ \begin {align*} M &= \dfrac {90A - 81}{A+18} + 10 \\&= \dfrac {90A + 90 \times 18 - 90 \times 18 - 81}{A + 18} + 10 \\&= 90 - \dfrac {90 \times 18 + 81}{A + 18} + 10. \]Aha! We want to minimize this expression, so we want to maximize $\dfrac{1}{A+18}$. And, to do this, we need to minimize $A$, since $A$ is in the denominator. This is reverse logic from what we did with $B$ and $C$. Since $A$ cannot be $0$, it must be $1$. We got: $(A,B,C)=(1,9,9)$, so the $3$-digit number is $ \boxed {199} $. $ \blacksquare $ This solution was copied from my solution at Brilliant dot org.

Let $a$,$b$,$c$ denote the digits in the hundredth,tenth and ones place respectively. If we fix the sum of the digits $n$,then are trying to find a number with minimal $a$ such that $a+b+c=n$.If $n \ge 23$ then $b \le 9,c \le9 \Rightarrow a \ge 5 \Rightarrow \text{ratio} \ge 2$.For $n \le 22$ we try all the cases and get the answer as $\boxed{199}$

Let our number be $100a+10b+c$. Then we aim to minimize $\dfrac {100a+10b+c}{a+b+c}=1+\dfrac {9(11a+b)}{a+b+c}$. The expression is minimized if $c=9$. So now we attempt to minimize $\dfrac {11a+b}{a+b+9}=\dfrac {a+b+9+10a-9}{a+b+9}=1+\dfrac {10a-9}{a+b+9}$ By previous logic, this is again minimized if $b=9$ as the numerator is independent of $b$. So finally we need to minimize $\dfrac {10a-9}{a+18}=\dfrac {10(a+18)-189}{a+18}=10-\dfrac {189}{a+18}$. This is minimized if $a$ is minimized i.e. $a=1$. This gives our answer $[b]199[/b]$.

SOLUTION Let our number be $100a+10b+c$. Then, we seek to mimize $r=\frac{100a+10b+c}{a+b+c}=1+\frac{99a+9b}{a+b+c}$. Hence, $c=9$. Replacing it, we have $r=1+\frac{99a+9b)}{a+b+9}=10+\frac{90x-81}{x+y+9}$. It follows that $y=9$. Furthermore, $r=10+\frac{90x-81}{x+18}=100-\frac{90 \cdot 18}{18+x}$. Hence, we want to minimize $x$ by letting it equal to $1$. Our final number is $\boxed{199}$. NOTES: This is quite an easy TST problem .