A

Let the mass of liquid A be $m_A$ and that of liquid B be $m_B$. Since they are equal: $m_A = m_B = m$.
The specific heat of liquid A is $c_A$ and that of liquid B is $c_B$.
According to the principle of conservation of energy, the heat lost by the hotter liquid will equal the heat gained by the cooler liquid. We have:

Heat lost by A:
$Q_A = m c_A (T_A - T_f)$
$Q_A = m c_A (30 - 26)$
$Q_A = m c_A imes 4$

Heat gained by B:
$Q_B = m c_B (T_f - T_B)$
$Q_B = m c_B (26 - 20)$
$Q_B = m c_B imes 6$

Setting the heat lost equal to the heat gained:
$m c_A imes 4 = m c_B imes 6$
Since mass cancels out, we get:
$4c_A = 6c_B$
Rearranging gives us the ratio:
$\frac{c_A}{c_B} = \frac{6}{4} = \frac{3}{2}$

Hence, the ratio of specific heats $\frac{m}{n} = \frac{3}{2}$. Therefore, $m = 3$ and $n = 2$.
So, $m + n = 3 + 2 = 5$.
Therefore, the minimum value of $m + n$ is 5.