A quadruple $$(a, b, c, d)$$ of distinct integers is said to be balanced if $$a+c=b+d$$. Let $$S$$ be any set of quadruples $$(a, b, c, d)$$ where $$1\le a<b<d<c\le 20$$ and where the cardinality of $$S$$ is $$4411$$. Find the least number of balanced quadruples in $$S$$.

Let the four integers be written in increasing order as $$x_1\lt x_2\lt x_3\lt x_4$$.
With this notation we have $$a=x_1,\; b=x_2,\; d=x_3,\; c=x_4$$ (because the required order is $$a\lt b\lt d\lt c$$), and every choice of four distinct integers from $$1\!-\!20$$ gives exactly one admissible quadruple.

Total number of admissible quadruples
    $$= \binom{20}{4}=4845$$.

A quadruple is balanced when $$a+c=b+d$$, i.e. when $$x_1+x_4=x_2+x_3$$.
Re-writing the equality gives $$x_4-x_3 = x_2-x_1$$.
Thus the outer two gaps must be equal. Set that common gap to $$d\,(d\ge 1)$$:

Case d fixed
Choose $$x_1$$ and $$x_3$$ such that

$$\begin{aligned} x_2 &= x_1+d, & x_4 &= x_3+d,\\ 1 &\le x_1, & x_4 &\le 20,\\ x_2&=x_1+d &\lt &x_3,\\ x_3&\lt x_4=x_3+d. \end{aligned}$$

The inequalities reduce to

$$\begin{cases} 1\le x_1\le 19-2d,\\[4pt] x_1+d+1\le x_3\le 20-d. \end{cases}$$

For a given $$d$$ the number of possible $$x_3$$ for each $$x_1$$ is
$$\bigl(20-d\bigr)-\bigl(x_1+d+1\bigr)+1 = 20-2d-x_1.$$

Hence the number of balanced quadruples for that $$d$$ is

$$ N_d=\sum_{x_1=1}^{19-2d}\bigl(20-2d-x_1\bigr) =(19-2d)\bigl(20-2d\bigr)-\frac{(19-2d)(20-2d)}{2} =\frac{(19-2d)(18-2d)}{2}. $$

Valid $$d$$ values satisfy $$19-2d\ge 1\;\Longrightarrow\;d\le 9$$. Computing $$N_d$$:

$$\begin{array}{c|ccccccccc} d & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9\\ \hline N_d & 153 & 120 & 91 & 66 & 45 & 28 & 15 & 6 & 1 \end{array}$$

Total balanced quadruples
$$N=\sum_{d=1}^{9} N_d = 153+120+91+66+45+28+15+6+1 = 525.$$

Therefore

$$\text{unbalanced quadruples}=4845-525=4320.$$

We must choose $$4411$$ quadruples for the set $$S$$. Even if we take all unbalanced ones (4320 of them), we still need
$$4411-4320=91$$ more, and every extra choice will necessarily be balanced.

Hence the least possible number of balanced quadruples in any such set $$S$$ is $$91$$.

Answer: 91