i) How many integers $n$ are there such that $n$ is divisible by $9$ and $n+1$ is divisible by $25$?
ii) How many integers $n$ are there such that $n$ is divisible by $21$ and $n+1$ is divisible by $165$?
iii) How many integers $n$ are there such that $n$ is divisible by $9, n + 1$ is divisible by $25$, and $n + 2$ is divisible by $4$?

### Part (i)
We need to find the integers \( n \) such that \( n \) is divisible by \( 9 \) and \( n+1 \) is divisible by \( 25 \).

1. Let \( n = 9a \) for some integer \( a \).
2. Since \( n+1 \) is divisible by \( 25 \), we have \( 9a + 1 = 25b \) for some integer \( b \).
3. Rearrange the equation: 
   \[
   9a = 25b - 1
   \]
4. We need \( 9a \) to be an integer, so \( 25b - 1 \) must be divisible by \( 9 \). This implies:
   \[
   25b - 1 \equiv 0 \pmod{9}
   \]
5. Simplify the congruence:
   \[
   25b \equiv 1 \pmod{9}
   \]
6. Since \( 25 \equiv 7 \pmod{9} \), we have:
   \[
   7b \equiv 1 \pmod{9}
   \]
7. Find the multiplicative inverse of \( 7 \) modulo \( 9 \). We need \( x \) such that:
   \[
   7x \equiv 1 \pmod{9}
   \]
   By trial, we find \( x = 4 \) works because:
   \[
   7 \cdot 4 = 28 \equiv 1 \pmod{9}
   \]
8. Thus, \( b \equiv 4 \pmod{9} \), so \( b = 9k + 4 \) for some integer \( k \).
9. Substitute \( b \) back into the equation \( n = 9a \):
   \[
   9a = 25(9k + 4) - 1 = 225k + 100 - 1 = 225k + 99
   \]
10. Therefore, \( n = 225k + 99 \).

### Part (ii)
We need to find the integers \( n \) such that \( n \) is divisible by \( 21 \) and \( n+1 \) is divisible by \( 165 \).

1. Let \( n = 21a \) for some integer \( a \).
2. Since \( n+1 \) is divisible by \( 165 \), we have \( 21a + 1 = 165b \) for some integer \( b \).
3. Rearrange the equation:
   \[
   21a = 165b - 1
   \]
4. We need \( 21a \) to be an integer, so \( 165b - 1 \) must be divisible by \( 21 \). This implies:
   \[
   165b - 1 \equiv 0 \pmod{21}
   \]
5. Simplify the congruence:
   \[
   165b \equiv 1 \pmod{21}
   \]
6. Since \( 165 \equiv 18 \pmod{21} \), we have:
   \[
   18b \equiv 1 \pmod{21}
   \]
7. Find the multiplicative inverse of \( 18 \) modulo \( 21 \). We need \( x \) such that:
   \[
   18x \equiv 1 \pmod{21}
   \]
   By trial, we find \( x = 10 \) works because:
   \[
   18 \cdot 10 = 180 \equiv 1 \pmod{21}
   \]
8. Thus, \( b \equiv 10 \pmod{21} \), so \( b = 21k + 10 \) for some integer \( k \).
9. Substitute \( b \) back into the equation \( n = 21a \):
   \[
   21a = 165(21k + 10) - 1 = 3465k + 1650 - 1 = 3465k + 1649
   \]
10. Therefore, \( n = 3465k + 1649 \).

### Part (iii)
We need to find the integers \( n \) such that \( n \) is divisible by \( 9 \), \( n+1 \) is divisible by \( 25 \), and \( n+2 \) is divisible by \( 4 \).

1. From part (i), we have \( n = 225k + 99 \).
2. We need \( n+2 \) to be divisible by \( 4 \):
   \[
   225k + 99 + 2 = 225k + 101
   \]
3. Check the divisibility:
   \[
   225k + 101 \equiv 0 \pmod{4}
   \]
4. Since \( 225 \equiv 1 \pmod{4} \), we have:
   \[
   k + 101 \equiv 0 \pmod{4}
   \]
5. Simplify the congruence:
   \[
   k \equiv -101 \pmod{4}
   \]
6. Since \( -101 \equiv 3 \pmod{4} \), we have:
   \[
   k \equiv 3 \pmod{4}
   \]
7. Thus, \( k = 4m + 3 \) for some integer \( m \).
8. Substitute \( k \) back into the equation \( n = 225k + 99 \):
   \[
   n = 225(4m + 3) + 99 = 900m + 675 + 99 = 900m + 774
   \]
9. Therefore, \( n = 900m + 774 \).

i) The final answer is \( \boxed{ n = 225k + 99 } \).