Determine all three-digit numbers $N$ having the property that $N$ is divisible by 11, and $\dfrac{N}{11}$ is equal to the sum of the squares of the digits of $N$.
To determine all three-digit numbers \( N \) such that \( N \) is divisible by 11 and \(\frac{N}{11}\) is equal to the sum of the squares of the digits of \( N \), we can proceed as follows:

1. **Express \( N \) in terms of its digits:**
   Let \( N = 100a + 10b + c \), where \( a, b, c \) are the digits of \( N \) and \( a \neq 0 \) since \( N \) is a three-digit number.

2. **Condition for divisibility by 11:**
   A number is divisible by 11 if the alternating sum of its digits is divisible by 11. Therefore, we need:
   \[
   a - b + c \equiv 0 \pmod{11}
   \]

3. **Condition for the sum of squares:**
   We are given that:
   \[
   \frac{N}{11} = a^2 + b^2 + c^2
   \]
   Thus:
   \[
   N = 11(a^2 + b^2 + c^2)
   \]

4. **Combine the conditions:**
   Substitute \( N = 100a + 10b + c \) into the equation \( N = 11(a^2 + b^2 + c^2) \):
   \[
   100a + 10b + c = 11(a^2 + b^2 + c^2)
   \]

5. **Check possible values of \( a \):**
   Since \( N \) is a three-digit number, \( a \) ranges from 1 to 9. We will check each case to see if there are valid solutions.

### Case 1: \( a = 1 \)
   \[
   100(1) + 10b + c = 11(1^2 + b^2 + c^2)
   \]
   \[
   100 + 10b + c = 11(1 + b^2 + c^2)
   \]
   \[
   100 + 10b + c = 11 + 11b^2 + 11c^2
   \]
   \[
   89 + 10b + c = 11b^2 + 11c^2
   \]
   Testing values of \( b \) and \( c \) within the range 0 to 9, we find no valid solutions.

### Case 2: \( a = 2 \)
   \[
   100(2) + 10b + c = 11(2^2 + b^2 + c^2)
   \]
   \[
   200 + 10b + c = 11(4 + b^2 + c^2)
   \]
   \[
   200 + 10b + c = 44 + 11b^2 + 11c^2
   \]
   \[
   156 + 10b + c = 11b^2 + 11c^2
   \]
   Testing values of \( b \) and \( c \) within the range 0 to 9, we find no valid solutions.

### Case 3: \( a = 3 \)
   \[
   100(3) + 10b + c = 11(3^2 + b^2 + c^2)
   \]
   \[
   300 + 10b + c = 11(9 + b^2 + c^2)
   \]
   \[
   300 + 10b + c = 99 + 11b^2 + 11c^2
   \]
   \[
   201 + 10b + c = 11b^2 + 11c^2
   \]
   Testing values of \( b \) and \( c \) within the range 0 to 9, we find no valid solutions.

### Case 4: \( a = 4 \)
   \[
   100(4) + 10b + c = 11(4^2 + b^2 + c^2)
   \]
   \[
   400 + 10b + c = 11(16 + b^2 + c^2)
   \]
   \[
   400 + 10b + c = 176 + 11b^2 + 11c^2
   \]
   \[
   224 + 10b + c = 11b^2 + 11c^2
   \]
   Testing values of \( b \) and \( c \) within the range 0 to 9, we find no valid solutions.

### Case 5: \( a = 5 \)
   \[
   100(5) + 10b + c = 11(5^2 + b^2 + c^2)
   \]
   \[
   500 + 10b + c = 11(25 + b^2 + c^2)
   \]
   \[
   500 + 10b + c = 275 + 11b^2 + 11c^2
   \]
   \[
   225 + 10b + c = 11b^2 + 11c^2
   \]
   Testing values of \( b \) and \( c \) within the range 0 to 9, we find that \( b = 5 \) and \( c = 0 \) works:
   \[
   225 + 10(5) + 0 = 11(5^2 + 0^2)
   \]
   \[
   275 = 275
   \]
   Thus, \( N = 550 \) is a solution.

### Case 6: \( a = 6 \)
   \[
   100(6) + 10b + c = 11(6^2 + b^2 + c^2)
   \]
   \[
   600 + 10b + c = 11(36 + b^2 + c^2)
   \]
   \[
   600 + 10b + c = 396 + 11b^2 + 11c^2
   \]
   \[
   204 + 10b + c = 11b^2 + 11c^2
   \]
   Testing values of \( b \) and \( c \) within the range 0 to 9, we find no valid solutions.

### Case 7: \( a = 7 \)
   \[
   100(7) + 10b + c = 11(7^2 + b^2 + c^2)
   \]
   \[
   700 + 10b + c = 11(49 + b^2 + c^2)
   \]
   \[
   700 + 10b + c = 539 + 11b^2 + 11c^2
   \]
   \[
   161 + 10b + c = 11b^2 + 11c^2
   \]
   Testing values of \( b \) and \( c \) within the range 0 to 9, we find no valid solutions.

### Case 8: \( a = 8 \)
   \[
   100(8) + 10b + c = 11(8^2 + b^2 + c^2)
   \]
   \[
   800 + 10b + c = 11(64 + b^2 + c^2)
   \]
   \[
   800 + 10b + c = 704 + 11b^2 + 11c^2
   \]
   \[
   96 + 10b + c = 11b^2 + 11c^2
   \]
   Testing values of \( b \) and \( c \) within the range 0 to 9, we find that \( b = 0 \) and \( c = 3 \) works:
   \[
   96 + 10(0) + 3 = 11(0^2 + 3^2)
   \]
   \[
   99 = 99
   \]
   Thus, \( N = 803 \) is a solution.

### Case 9: \( a = 9 \)
   \[
   100(9) + 10b + c = 11(9^2 + b^2 + c^2)
   \]
   \[
   900 + 10b + c = 11(81 + b^2 + c^2)
   \]
   \[
   900 + 10b + c = 891 + 11b^2 + 11c^2
   \]
   \[
   9 + 10b + c = 11b^2 + 11c^2
   \]
   Testing values of \( b \) and \( c \) within the range 0 to 9, we find no valid solutions.

Thus, our only solutions are \( N = 550 \) and \( N = 803 \).

The final answer is \( \boxed{550} \) and \( \boxed{803} \).