12

See this image: https://imgur.com/a/fhwCVoI If 15 children ask for a bit more than $\frac{1}{16}$ of the cake, then each one of these 15 that gets a square will eat one of the blue candies, but there are only 9 blue candies, so only 9+3=12 children can be satisfied at most.

I claim we can give every child a square, except for the 6 children that wanted the most cake (they were greedy so they don't deserve it). Say the cake has area 1. Suppose child $i$ asked for $a_i$ cake, so that $a_1\geq a_2\geq \cdots\geq a_{18}$ and $a_1+a_2+\cdots+a_{18}=1$. We will try to fit a $4\times 3$ table of cells inside the cake, and cut a squares of areas $a_7, a_8, \dots, a_{18}$ out of the $12$ cells. The table will have dimensions as in this image: https://imgur.com/a/VpXO021 Out of the cell labelled $k$, it's possible to cut a square of area $a_k$. It remains to prove that this table fits inside the $1\times 1$ cake, in other words that \[\sqrt{a_7}+\sqrt{a_8}+\sqrt{a_9}\leq 1\quad \text{and} \quad \sqrt{a_7}+\sqrt{a_{10}}+\sqrt{a_{13}}+\sqrt{a_{16}}\leq 1\]By C-S it suffices to show that \[a_7+a_8+a_9\leq \frac{1}{3} \quad \text{and}\quad a_7+a_{10}+a_{13}+a_{16}\leq \frac{1}{4}\]But notice that $\sum_{i=1}^{9}a_i\leq 1$, and because the $a_i$ are decreasing we have $3(a_7+a_8+a_9)\leq \sum_{i=1}^9 a_i$, which proves the first inequality. The second follows from $\sum_{i=1}^{16}a_i\leq 1$ and \[a_7+a_{10}+a_{13}+a_{16}\leq a_{6}+a_{9}+a_{12}+a_{15}\leq a_5+a_8+a_{11}+a_{14}\leq a_1+a_2+a_3+a_4\]so that $4(a_7+a_{10}+a_{13}+a_{16})\leq \sum_{i=1}^{16}a_i$.$\blacksquare$