The minimum number is 5 4 portraits isn't enough We have that $ \frac {(a_1 + b_1)\cdot (a_2 + b_2)\cdot(a_3 + b_n3)\cdot(a_4 + b_4)}{a_1\cdot a_2\cdot a_3\cdot a_4} = \frac {5}{4}$ , with $ 1\leq b_1,b_2,b_3,b_4\leq 80$. The maximal value for $ \frac {a_1 + b_1}{a_1}$ is $ \frac {1680}{1600}$, for $ \frac {a_2 + b_2}{a_1}$ is $ \frac {1681}{1601}$ and so on. We easily see that $ \frac {(a_1 + b_1)\cdot (a_2 + b_2)\cdot(a_3 + b_n3)\cdot(a_4 + b_4)}{a_1\cdot a_2\cdot a_3\cdot a_4} < \frac {5}{4}$ 5 portraits is enough $ \frac {1680}{1601} = \frac {21}{20}$ is the maximal value for $ \frac {a_i + b_i}{a_i}$ and $ \frac {2008}{2007}$ is the minimal value. $ \frac {5}{4} = \frac {5}{4}\cdot\frac {5}{5} = \frac {25}{20} = \frac {21}{20}\cdot\frac {22}{21}\cdot\frac {23}{22}\cdot\frac {24}{23}\cdot\frac {25}{24} = c_1\cdot c_2\cdot c_3\cdot c_4\cdot c_5$. And hold $ \frac {1680}{1601}\leq c_1,c_2,c_3,c_4,c_5\leq\frac {2008}{2007}$. OK