(i) Easy to see by pigeonhole that if there are more than 7 black cells in a $7$x$2$, Julián lose. Let $C_i$ the number of black cells in the $i$'th column. (ii)If Julián can color the grid in such a way that it's impossible for Pilar to win, then: $C_1+C_2<8$ $C_3+C_4<8$ $C_5+C_6<8$ Adding:$C_1+C_2+C_3+C_4+C_5+C_6<8 \cdot 3=24 \Rightarrow C_1+C_2+C_3+C_4+C_5+C_6+C_7=29<24+C_7 \Rightarrow 5<C_7$. Analogously, $5<C_1$. This implies that, because of (i), $C_2, C_6<2$, so $15 \le C_3+C_4+C_5 \le 17$. (iii) Using (i): $C_2+C_3<8$ $C_3+C_4<8$ Adding: $C_3+(C_2+C_3+C_4)<16$. As $C_3$ is a non-negative value, $C_2+C_3+C_4=15 \Rightarrow C_3<1 \Rightarrow C_3=0$, but then that implies that $C_2+C_4=15$ and they are $7$x$2$. So $C_2+C_4<15$. Contradiction. Julián can't win.