14th Turkish Mathematical Olympiad

$t=2006/14$. If every student is acquainted with every teacher, then all relevant ratios are $2006/14$. This means $t\le 2006/14$. Now we will show that $t\ge 2006/14$. For $1\le i\le 14$, let $a_i$ denote the number of students who are acquainted with the $i$\text{th} teacher, and for $1\le j\le 2006$, let $b_j$ denote the number of teachers who are acquainted with the $j$\text{th} student. Let $T$ be the set of ordered pairs $(i,j)$ of teachers and students who are acquainted with each other. Observe that with this notation we have $|\{j:(i,j)\in T\}|=a_i\ge 0$ and $|\{i:(i,j)\in T\}|=b_j\ge 1$. Assume that $(i_o,j_o)\in T$ satisfies the condition $a_{i_o}/b_{j_o}\ge a_i/b_j$ for all $(i,j)\in T$. Adding the inequalities $a_{i_o}/b_{j_o}\cdot 1/a_i\ge 1/b_j$ for all $(i,j)\in T$ we obtain the inequality $$\frac{a_{i_o}}{b_{j_o}}\sum _{(i,j)\in T}\frac{1}{a_i}\ge \sum _{(i,j)\in T}\frac{1}{b_j}.$$ Since $$\sum _{(i,j)\in T}\frac{1}{a_i}=\sum _{i=1}^{14}\sum _{\{j:(i,j)\in T\}}\frac{1}{a_i}=\sum _{\{i:a_i\ne 0\}}\frac{a_i}{a_i}=\sum _{\{i:a_i\ne 0\}}1\le 14,$$ and similarly, $$\sum _{(i,j)\in T}\frac{1}{b_j}=\sum _{j=1}^{2006}\sum _{\{i:(i,j)\in T\}}\frac{1}{b_j}=\sum _{j=1}^{2006}\frac{b_j}{b_j}=\sum _{j=1}^{2006}1=2006,$$ we deduce that $a_{i_o}/b_{j_o}\ge 2006/14$, and hence $t\ge 2006/14$.