For every positive integer $n$, form the number $n/s(n)$, where $s(n)$ is the sum of digits of $n$ in base 10. Determine the minimum value of $n/s(n)$ in each of the following cases:

(i) $10 \leq n \leq 99$
(ii) $100 \leq n \leq 999$
(iii) $1000 \leq n \leq 9999$
(iv) $10000 \leq n \leq 99999$
To determine the minimum value of \( \frac{n}{s(n)} \) for each given range of \( n \), we need to analyze the expression and find the values of \( n \) that minimize this ratio. Here, \( s(n) \) represents the sum of the digits of \( n \).

### Case (i): \( 10 \leq n \leq 99 \)

1. Let \( n = 10a + b \), where \( a \) and \( b \) are digits (i.e., \( 1 \leq a \leq 9 \) and \( 0 \leq b \leq 9 \)).
2. The sum of the digits \( s(n) = a + b \).
3. The ratio \( \frac{n}{s(n)} = \frac{10a + b}{a + b} \).

To minimize \( \frac{10a + b}{a + b} \), we need to minimize \( a \) and maximize \( b \):
\[ \frac{10a + b}{a + b} = 1 + \frac{9a}{a + b} \]

- For \( a = 1 \) and \( b = 9 \):
\[ \frac{10 \cdot 1 + 9}{1 + 9} = \frac{19}{10} = 1.9 \]

Thus, the minimum value of \( \frac{n}{s(n)} \) for \( 10 \leq n \leq 99 \) is \( \boxed{\frac{19}{10}} \).

### Case (ii): \( 100 \leq n \leq 999 \)

1. Let \( n = 100a + 10b + c \), where \( a, b, c \) are digits (i.e., \( 1 \leq a \leq 9 \) and \( 0 \leq b, c \leq 9 \)).
2. The sum of the digits \( s(n) = a + b + c \).
3. The ratio \( \frac{n}{s(n)} = \frac{100a + 10b + c}{a + b + c} \).

To minimize \( \frac{100a + 10b + c}{a + b + c} \), we need to minimize \( a \) and \( b \) and maximize \( c \):
\[ \frac{100a + 10b + c}{a + b + c} = 1 + \frac{99a + 9b}{a + b + c} \]

- For \( a = 1 \), \( b = 1 \), and \( c = 9 \):
\[ \frac{100 \cdot 1 + 10 \cdot 1 + 9}{1 + 1 + 9} = \frac{119}{11} \approx 10.818 \]

Thus, the minimum value of \( \frac{n}{s(n)} \) for \( 100 \leq n \leq 999 \) is \( \boxed{\frac{119}{11}} \).

### Case (iii): \( 1000 \leq n \leq 9999 \)

1. Let \( n = 1000a + 100b + 10c + d \), where \( a, b, c, d \) are digits (i.e., \( 1 \leq a \leq 9 \) and \( 0 \leq b, c, d \leq 9 \)).
2. The sum of the digits \( s(n) = a + b + c + d \).
3. The ratio \( \frac{n}{s(n)} = \frac{1000a + 100b + 10c + d}{a + b + c + d} \).

To minimize \( \frac{1000a + 100b + 10c + d}{a + b + c + d} \), we need to minimize \( a \) and \( b \) and \( c \) and maximize \( d \):
\[ \frac{1000a + 100b + 10c + d}{a + b + c + d} = 1 + \frac{999a + 99b + 9c}{a + b + c + d} \]

- For \( a = 1 \), \( b = 1 \), \( c = 1 \), and \( d = 9 \):
\[ \frac{1000 \cdot 1 + 100 \cdot 1 + 10 \cdot 1 + 9}{1 + 1 + 1 + 9} = \frac{1119}{12} \approx 93.25 \]

Thus, the minimum value of \( \frac{n}{s(n)} \) for \( 1000 \leq n \leq 9999 \) is \( \boxed{\frac{1119}{12}} \).

### Case (iv): \( 10000 \leq n \leq 99999 \)

1. Let \( n = 10000a + 1000b + 100c + 10d + e \), where \( a, b, c, d, e \) are digits (i.e., \( 1 \leq a \leq 9 \) and \( 0 \leq b, c, d, e \leq 9 \)).
2. The sum of the digits \( s(n) = a + b + c + d + e \).
3. The ratio \( \frac{n}{s(n)} = \frac{10000a + 1000b + 100c + 10d + e}{a + b + c + d + e} \).

To minimize \( \frac{10000a + 1000b + 100c + 10d + e}{a + b + c + d + e} \), we need to minimize \( a \) and \( b \) and \( c \) and \( d \) and maximize \( e \):
\[ \frac{10000a + 1000b + 100c + 10d + e}{a + b + c + d + e} = 1 + \frac{9999a + 999b + 99c + 9d}{a + b + c + d + e} \]

- For \( a = 1 \), \( b = 1 \), \( c = 1 \), \( d = 1 \), and \( e = 9 \):
\[ \frac{10000 \cdot 1 + 1000 \cdot 1 + 100 \cdot 1 + 10 \cdot 1 + 9}{1 + 1 + 1 + 1 + 9} = \frac{11119}{13} \approx 855.31 \]

Thus, the minimum value of \( \frac{n}{s(n)} \) for \( 10000 \leq n \leq 99999 \) is \( \boxed{\frac{11119}{13}} \).