Answer: $m = \large\frac{4m_1m_2}{m_1+m_2}$

In the system on the right, we have $mg - T = ma' \rightarrow T = m (g-a')$

Net force in the direction of motion of $m_1$ is $F_1 = m_1g - T/2$ which is equal to $m_1(a'-a)$ as per Newton's second law.

Net force in the direction of motion of $m_1$ is $F_2 = m_2g - T/2$ which is equal to $m_2(a'+a)$ as per Newton's second law.

Here $a$ represents the acceleration of $m_1$ and $m_2$.

Let's define a new variable $g'$ such that $g' = g-a$. We can rewrite the equations as follows:

$(i)\;\; \quad mg' = T$

$(ii)\; \quad m_1g' = T/2 - m_1a$

$(iii) \quad m_2g' = T/2 +m_2a$

Eliminating $a$ from $(ii)$ and $(iii)$ and solving for $T$, gives us:

$\Rightarrow T = \large\frac{4m_1m_2}{m_1+m_2}$$g'$

Substituting in $(i)$, we get: $m = \large\frac{4m_1m_2}{m_1+m_2}$

The two-mass system in the left may be equivalently treated as a mass m, given by this equation above, as far as the upper string is concerned