Since $a b \cdot c d=a c \cdot b d=a d \cdot b c$, the largest sum among $a b+c d, a c+b d, a d+b c$ will be the one with the largest difference between the two quantities, so $a b+c d=100, a c+b d=70, a d+b c=40$. Consider the sum of each pair of equations, which gives $(a+b)(c+d)=110,(a+c)(b+d)=140,(a+$ $d)(b+c)=170$. Since each of these are pairs summing to $S=a+b+c+d$, by looking at the discriminant of the quadratics with roots $a+b$ and $c+d, a+c$ and $b+d$, and $a+d$ and $b+c$, we have that $S^{2}-680, S^{2}-560, S^{2}-440$ must be perfect squares. Therefore, we need to find all arithmetic progressions of three squares with common difference 120. The equation $x^{2}-y^{2}=120$ has $(31,29),(17,13),(13,7),(11,1)$ as a solution, and so the only possibility is $49,169,289$. This implies that $S^{2}=729 \Longrightarrow S=27$. We now need to verify this works. Note that this implies $\{a+b, c+d\}=\{5,22\},\{a+c, b+d\}=$ $\{7,20\},\{a+d, b+c\}=\{10,17\}$. Therefore, $a+b=5, a+c=7$. This means that $b+c$ is even, so $b+c=10$. This gives us $(a, b, c, d)=(1,4,6,16)$ is the only possibility, as desired.