Three of the $16$ squares from a $4 \times 4$ grid of squares are selected at random. The probability that at least one corner square of the grid is selected is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$.

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To solve this problem, we will use the concept of complementary probability. The complementary probability approach involves finding the probability of the complement of the desired event and then subtracting it from 1.

1. **Total number of ways to select 3 squares from 16 squares:**
   \[
   \binom{16}{3} = \frac{16 \times 15 \times 14}{3 \times 2 \times 1} = 560
   \]

2. **Number of ways to select 3 squares such that none of them is a corner square:**
   - There are 4 corner squares in a \(4 \times 4\) grid.
   - Therefore, there are \(16 - 4 = 12\) non-corner squares.
   - The number of ways to select 3 squares from these 12 non-corner squares is:
     \[
     \binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220
     \]

3. **Probability that none of the selected squares is a corner square:**
   \[
   \frac{\binom{12}{3}}{\binom{16}{3}} = \frac{220}{560} = \frac{11}{28}
   \]

4. **Probability that at least one corner square is selected:**
   \[
   1 - \frac{11}{28} = \frac{28}{28} - \frac{11}{28} = \frac{17}{28}
   \]

Thus, the probability that at least one corner square is selected is \(\frac{17}{28}\).

The final answer is \(17 + 28 = \boxed{45}\).