Problem 1236

1. Understanding the problem: We need to find the least number of balanced quadruples $(a, b, c, d)$ in a set $\mathcal{S}$ where $1 \leq a < b < d < c \leq 20$ and the cardinality of $\mathcal{S}$ is 4411. A quadruple $(a, b, c, d)$ is balanced if $a + c = b + d$.

2. Total number of quadruples: First, we calculate the total number of quadruples $(a, b, c, d)$ where $1 \leq a < b < d < c \leq 20$. This is given by the binomial coefficient:
$$\binom{20}{4} = \frac{20 \times 19 \times 18 \times 17}{4 \times 3 \times 2 \times 1} = 4845$$

3. Number of unbalanced quadruples: Since the cardinality of $\mathcal{S}$ is 4411, the number of unbalanced quadruples is:
$$4845 - 4411 = 434$$

4. Balanced quadruples: We need to find the least number of balanced quadruples. We start by calculating the possible values of $a + c$. The minimum value of $a + c$ is $1 + 4 = 5$ (since $a < b < d < c$) and the maximum value is $17 + 20 = 37$.

5. **Counting balanced quadruples for each $a + c$**:
- For $a + c = k$, we need to count the number of ways to choose $a$ and $c$ such that $a < c$ and $a + c = k$.
- For $k = 5$, the only possibility is $(1, 4)$.
- For $k = 6$, the possibilities are $(1, 5)$ and $(2, 4)$.
- For $k = 7$, the possibilities are $(1, 6)$, $(2, 5)$, and $(3, 4)$.
- This pattern continues until $k = 20$, where the possibilities are $(10, 10)$.

6. Summing the number of balanced quadruples:
- For $k = 5$ to $k = 20$, the number of pairs $(a, c)$ is given by $\binom{k-1}{2}$.
- For $k = 21$, the number of pairs $(a, c)$ is $\binom{9}{2}$.
- For $k = 22$ to $k = 37$, the number of pairs $(a, c)$ is symmetric to the pairs for $k = 5$ to $k = 20$.

7. Calculating the total number of balanced quadruples:
$$2 \cdot \left(\binom{2}{2} + \binom{3}{2} + \cdots + \binom{9}{2}\right) + \binom{10}{2} = 2 \cdot \left(1 + 3 + 6 + 10 + 15 + 21 + 28 + 36\right) + 45 = 2 \cdot 120 + 45 = 285$$
- For $a + c = 21$, we have $\binom{9}{2} = 36$.

8. Total balanced quadruples:
$$285 + 36 = 321$$

9. Least number of balanced quadruples:
- Since we have 434 unbalanced quadruples, the least number of balanced quadruples is:
$$525 - 434 = 91$$

The final answer is $\boxed{91}$