Three numbers $N,n,r$ are such that the digits of $N,n,r$ taken together are formed by $1,2,3,4,5,6,7,8,9$ without repetition. 
If $N = n^2 - r$, find all possible combinations of $N,n,r$.
Given three numbers \( N, n, r \) such that the digits of \( N, n, r \) taken together are formed by \( 1, 2, 3, 4, 5, 6, 7, 8, 9 \) without repetition. We are also given that \( N = n^2 - r \). We need to find all possible combinations of \( N, n, r \).

First, we note that the sum of the digits \( 1, 2, 3, 4, 5, 6, 7, 8, 9 \) is:
\[
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45
\]
Since \( N, n, r \) use all these digits without repetition, we have:
\[
N + n + r = 45
\]
Given \( N = n^2 - r \), we can substitute \( r \) to get:
\[
N + n + (n^2 - N) = 45 \implies n^2 + n = 45 \implies n^2 + n - 45 = 0
\]
Solving this quadratic equation:
\[
n = \frac{-1 \pm \sqrt{1 + 4 \cdot 45}}{2} = \frac{-1 \pm \sqrt{181}}{2}
\]
Since \(\sqrt{181}\) is not an integer, \( n \) must be an integer. Therefore, we need to check possible values of \( N \) that are divisible by 9 (since the sum of the digits is 45, which is divisible by 9).

Possible values of \( N \) that are two-digit numbers and divisible by 9 are:
\[
N = 36, 45, 54, 63, 72
\]

We will check each case:

1. **Case \( N = 36 \)**:
   \[
   n^2 - r = 36 \implies r = n^2 - 36
   \]
   We need \( n \) and \( r \) to use the remaining digits \( 1, 2, 3, 4, 5, 6, 7, 8, 9 \) without repetition. Checking possible values of \( n \):
   \[
   n = 6 \implies r = 6^2 - 36 = 0 \quad (\text{not valid since } r \text{ must be a positive integer})
   \]
   No valid solution for \( N = 36 \).

2. **Case \( N = 45 \)**:
   \[
   n^2 - r = 45 \implies r = n^2 - 45
   \]
   Checking possible values of \( n \):
   \[
   n = 7 \implies r = 7^2 - 45 = 4 \quad (\text{valid})
   \]
   Digits used: \( N = 45 \), \( n = 7 \), \( r = 4 \). Remaining digits: \( 1, 2, 3, 6, 8, 9 \).

3. **Case \( N = 54 \)**:
   \[
   n^2 - r = 54 \implies r = n^2 - 54
   \]
   Checking possible values of \( n \):
   \[
   n = 8 \implies r = 8^2 - 54 = 10 \quad (\text{not valid since } r \text{ must be a single digit})
   \]
   No valid solution for \( N = 54 \).

4. **Case \( N = 63 \)**:
   \[
   n^2 - r = 63 \implies r = n^2 - 63
   \]
   Checking possible values of \( n \):
   \[
   n = 8 \implies r = 8^2 - 63 = 1 \quad (\text{valid})
   \]
   Digits used: \( N = 63 \), \( n = 8 \), \( r = 1 \). Remaining digits: \( 2, 3, 4, 5, 7, 9 \).

5. **Case \( N = 72 \)**:
   \[
   n^2 - r = 72 \implies r = n^2 - 72
   \]
   Checking possible values of \( n \):
   \[
   n = 9 \implies r = 9^2 - 72 = 9 \quad (\text{valid})
   \]
   Digits used: \( N = 72 \), \( n = 9 \), \( r = 9 \). Remaining digits: \( 1, 2, 3, 4, 5, 6, 7, 8 \).

Thus, the valid combinations are:
\[
(N, n, r) = (45, 7, 4), (63, 8, 1), (72, 9, 9)
\]

The final answer is \( \boxed{ (45, 7, 4), (63, 8, 1), (72, 9, 9) } \)