How many more men than women were there in the beginning? 
Let men=m, women=w, children=c
From the statement,
m=1.25w
0.8w=c
Children left resulted in equal number of men and women, women=122 at the end.
Can I assumed that the total number of people is 122+122=244?
m+w+c=244
1.25w+w+0.8w=244
3.05w=244
w=80
m=100
The number of men more than women
m-w=100-80=20
Is it the correct way to solve this problem?
 A: I think this question would benefit from the addition of indices to the variables, since the values are changing. The indices will be i for initial and f for final.
$$m_i=(1+0.25)w_i=1.25w_i$$
$$c_i=(1-0.20)w_i=0.8w_i$$
$$c_f=0$$
$$m_f+w_f-(m_i+w_i)=c_i-c_f$$
$$m_f=w_f$$
$$w_f=122$$
$$\text{Solve for: }m_i-w_i$$
Here we have six independent equations and six unknowns. We could use matrix math to solve this, or it is a simple enough problem to take the steps you did. Let's go the latter route.
$$w_f=122$$
$$m_f=w_f=122$$
$$m_f+w_f-(m_i+w_i)=122+122-(1.25w_i+w_i)=0.8w_i-0=c_i-c_f$$
$$244-2.25w_i=0.8w_i$$
$$244=3.05w_i$$
$$w_i=80$$
$$m_i=1.25w_i=100$$
$$m_i-w_i=20$$
There were 20 more men than women at the beginning of the book fair.
A: Here's an alternative working, for your reference.

*

*Summarising the given information:
$$M_1:W_1:C=125:100:80=25:20:16;\\\text{The children are all
replaced with equally many adults;}\\M_2:W_2=1:1;\\W_2=122.$$

*Let the number of children be $u.$ Then
$$\frac{25}{16}u+\frac{20}{16}u+u=122+122\\u=64;$$ therefore, at the
beginning, there were
$\displaystyle\frac{25}{16}u-\frac{20}{16}u=20$ more men than women.

