Find the sum of all four-digit integers whose digits are a rearrangement of the digits $1$, $2$, $3$, $4$, such as $1234$, $1432$, or $3124$.
1. **Calculate the sum of the digits:**
   \[
   1 + 2 + 3 + 4 = 10
   \]

2. **Determine the number of four-digit integers that can be formed using the digits \(1, 2, 3, 4\):**
   Since we are rearranging 4 unique digits, the number of permutations is:
   \[
   4! = 24
   \]

3. **Calculate the contribution of each digit to each place value (units, tens, hundreds, thousands):**
   Each digit appears in each place value exactly \(\frac{24}{4} = 6\) times because there are 24 permutations and 4 positions.

4. **Sum the contributions for each place value:**
   - **Thousands place:**
     \[
     6 \times (1 + 2 + 3 + 4) \times 1000 = 6 \times 10 \times 1000 = 60000
     \]
   - **Hundreds place:**
     \[
     6 \times (1 + 2 + 3 + 4) \times 100 = 6 \times 10 \times 100 = 6000
     \]
   - **Tens place:**
     \[
     6 \times (1 + 2 + 3 + 4) \times 10 = 6 \times 10 \times 10 = 600
     \]
   - **Units place:**
     \[
     6 \times (1 + 2 + 3 + 4) \times 1 = 6 \times 10 \times 1 = 60
     \]

5. **Sum all the contributions:**
   \[
   60000 + 6000 + 600 + 60 = 66660
   \]

The final answer is \(\boxed{66660}\).