$5/2$

Split the square into two congruent rectangles, the sum of written numbers is $2+1/2=5/2$.

Suppose the square is $A \times A$, blue rectangles are $b_i \times c_i$, red rectangles are $r_j \times s_j$. ( $ \text{horizontal length} \times \text{vertical length}$.) The area condition implies $\sum b_ic_i = \sum r_js_j = A^2/2$. The crucial claim is Claim. $\sum c_i \ge A$ or $\sum r_j \ge A$. Proof. Suppose that $\sum c_i < A$. Then there exists a horizontal line, which doesn't pass through any red rectangles. Therefore, the sum of the horizontal lengths of the blue rectangles it passes through is $A$, so $\sum r_j \ge A$. WLOG suppose $\sum c_i \ge A$. Use C-S and note that $s_j \le A$, $$ \sum \frac{c_i}{b_i} + \sum \frac{r_j}{s_j} \ge \frac{\left( \sum c_i \right)^2}{\sum b_ic_i} + \sum \frac{r_js_j}{A^2} \ge 2+1/2=5/2. $$