Divide and Conquer

Let $N=\overline{abc}$ , where $a,b,c$ are the digits. It's pretty obvious that for round numbers $N=100, 200, . . . , 900$ we have $\frac{N}{a+b+c}=100$ . Moving on, when $N$ is not "round", then $b+c > 0$ and $a+b+c \geq a+1$ . And since the leading digit of $N$ is $a$ , then $N<(a+1)\cdot 100$ , and $\frac{N}{a+b+c}<\frac{(a+1)\cdot 100}{a+1}=100$

Thus, the largest possible value for that ratio is $\boxed{100}$ , which is only obtained for round numbers.

There are only $27$ sums of digits or cases of denominators. And each sum there is a maximum $\overline {abc}$ , therefore, a maximum ratio. The maximum ratios are tabulated below:

$\begin{matrix} a+b+c & max (\overline{abc}) & \frac {max (\overline{abc})}{a+b+c} \\ \hline \\ 1 & 100 & 100.0 \\ 2 & 200 & 100.0 \\ 3 & 300 & 100.0 \\ 4 & 400 & 100.0 \\ 5 & 500 & 100.0 \\ 6 & 600 & 100.0 \\ 7 & 700 & 100.0 \\ 8 & 800 & 100.0 \\ 9 & 900 & 100.0 \\ 10 & 910 & 91.0 \\ 11 & 920 & 83.6 \\ 12 & 930 & 77.5 \\ 13 & 940 & 72.3 \\ 14 & 950 & 67.9 \\ 15 & 960 & 64.0 \\ 16 & 970 & 60.6 \\ 17 & 980 & 57.6 \\ 18 & 990 & 55.0 \\ 19 & 991 & 52.2 \\ 20 & 992 & 49.6 \\ 21 & 993 & 47.3 \\ 22 & 994 & 45.2 \\ 23 & 995 & 43.3 \\ 24 & 996 & 41.5 \\ 25 & 997 & 39.9 \\ 26 & 998 & 38.4 \\ 27 & 999 & 37.0 \end{matrix}$

Therefore, the maximum ratio is $\boxed{100}$ .

Just find out the lowest sum possible made with integers: 0 to 9 and which 3 integers this is possible for; in this case it would be 0 + 0 + 1; thus the number can't be 001 but rather 100. Thus 100/ 1 = 100.

Computing tells that {100, 200, 300, 400, 500, 600, 700, 800, 900} gives a maximum = n x 100/ (n + 0 + 0) = 100