Let $ S $ denote the given sum. First, we apply the fact that for all real numbers $ x, $ $ \lfloor x \rfloor > x - 1 $. To see this, recall that any real number can be split up into its integer and fractional parts: \[x = \lfloor x \rfloor + \{x\}.\]The fractional part of a real number is always less than 1, so $ x < \lfloor x \rfloor + 1 $. Hence, $ \lfloor x \rfloor > x - 1 $. Then \begin{align*} \left\lfloor \frac{b + c + d}{a} \right\rfloor &> \frac{b + c + d}{a} - 1, \\ \left\lfloor \frac{a + c + d}{b} \right\rfloor &> \frac{a + c + d}{b} - 1, \\ \left\lfloor \frac{a + b + d}{c} \right\rfloor &> \frac{a + b + d}{c} - 1, \\ \left\lfloor \frac{a + b + c}{d} \right\rfloor &> \frac{a + b + c}{d} - 1.\end{align*}Adding these inequalities, we get \begin{align*} S &> \frac{b + c + d}{a} - 1 + \frac{a + c + d}{b} - 1 + \frac{a + b + d}{c} - 1 + \frac{a + b + c}{d} - 1 \\ &= \frac{a}{b} + \frac{b}{a} + \frac{a}{c} + \frac{c}{a} + \frac{a}{d} + \frac{d}{a} + \frac{b}{c} + \frac{c}{b} + \frac{b}{d} + \frac{d}{b} + \frac{c}{d} + \frac{d}{c} - 4.\end{align*}By AM-GM, $ \frac{a}{b} + \frac{b}{a} \ge 2 $. The same applies to the other pairs of fractions, so $ S > 6 \cdot 2 - 4 = 8 $. As a sum of floors, $ S $ itself must be an integer, so $ S $ must be at least 9. When $ a = 4 $ and $ b = c = d = 5, $ $ S = 9 $. Therefore, the minimum value of $ S $ is $ \boxed{9} $.