Answer: $3^{28}$.
Let us numerate boards by $1$, $2$, $\ldots$, $29$ and for each student define a vector $(a_1, a_2, \ldots, a_{29})$, where $a_i$ is a number written by this student to the board number $i$. Since any two students wrote different numbers to at least one board, we get a set $V$ of $n$ different vectors each having $29$ components. For each given vector $v = (a_1, a_2, \ldots, a_{29})$ let us define $v_1 = (a_1+1, a_2+1, \ldots, a_{29}+1)$ and $v_2 = (a_1+2, a_2+2, \ldots, a_{29}+2)$ where all entries are considered in $(\bmod\ 3)$. Since any three students wrote the same number to at least one board, at most one of three vectors $v$, $v_1$ and $v_2$ can be present in the set $V$. Therefore, $V$ contains at most
$$\frac{3^{29}}{3} = 3^{28}$$
vectors. Now note that if first entries of all vectors of $V$ coincide and the remaining components span all possible $3^{28}$ combinations, problem conditions are satisfied. Thus, the maximal value of $n$ is $3^{28}$.