**Solution:**

Let's assume the original mixture contains x grams of NaCl and y grams of Na2CO3.

We are given that the mixture weighs 4 gm. So we can write the equation:
x + y = 4 ...(1)

Next, we are given that 250 ml of the solution required 50 ml of N/10 HCl for complete neutralization. This means that the moles of NaCl and Na2CO3 in the mixture react in a 1:1 ratio with the moles of HCl.

**Calculating the moles:**

The molar mass of NaCl is 58.44 g/mol and the molar mass of Na2CO3 is 105.99 g/mol.
So, the moles of NaCl in the mixture can be calculated as:
moles of NaCl = (x / 58.44) ...(2)

The moles of Na2CO3 in the mixture can be calculated as:
moles of Na2CO3 = (y / 105.99) ...(3)

Since the moles of NaCl and Na2CO3 are in a 1:1 ratio with HCl, the moles of HCl required for complete neutralization can be calculated as:
moles of HCl = (50 / 1000) * (1/10) = 0.005 ...(4)

**Setting up the equation:**

From equations (2) and (3), we know that:
moles of NaCl = moles of Na2CO3

So we can write:
(x / 58.44) = (y / 105.99)

Simplifying this equation, we get:
105.99x = 58.44y ...(5)

**Solving the equations:**

We now have two equations, equations (1) and (5), with two variables (x and y). We can solve these equations simultaneously to find the values of x and y.

Solving equations (1) and (5), we get:
105.99x - 58.44y = 0 ...(6)
x + y = 4 ...(7)

From equation (7), we can write:
x = 4 - y

Substituting this value in equation (6), we get:
105.99(4 - y) - 58.44y = 0

Simplifying this equation, we find:
423.96 - 105.99y - 58.44y = 0
164.55y = 423.96
y = 2.57

Substituting this value in equation (7), we get:
x + 2.57 = 4
x = 1.43

So, the original mixture contains 1.43 grams of NaCl and 2.57 grams of Na2CO3.

Therefore, the composition of the original mixture is:
NaCl:Na2CO3 = 1.43:2.57, which can be simplified to 1:1.8