jmc

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 $$\begin{array}{rl}\lfloor \frac{b+c+d}{a}\rfloor & >\frac{b+c+d}{a}-1,\\ \lfloor \frac{a+c+d}{b}\rfloor & >\frac{a+c+d}{b}-1,\\ \lfloor \frac{a+b+d}{c}\rfloor & >\frac{a+b+d}{c}-1,\\ \lfloor \frac{a+b+c}{d}\rfloor & >\frac{a+b+c}{d}-1.\end{array}$$Adding these inequalities, we get $$\begin{array}{rl}S& >\frac{b+c+d}{a}-1+\frac{a+c+d}{b}-1+\frac{a+b+d}{c}-1+\frac{a+b+c}{d}-1\\ & =\frac{a}{b}+\frac{b}{a}+\frac{a}{c}+\frac{c}{a}+\frac{a}{d}+\frac{d}{a}+\frac{b}{c}+\frac{c}{b}+\frac{b}{d}+\frac{d}{b}+\frac{c}{d}+\frac{d}{c}-4.\end{array}$$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 $\boxed{9}.$