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 $t \ge 2006/14$.