Let the amounts be:

• Moles of $NaOH = x$
• Moles of $Na_2CO_3 = y$

Reaction with $HCl$ up to phenolphthalein end point:

Phenolphthalein end point corresponds to:

• complete neutralization of $NaOH$
• conversion of $Na_2CO_3$ to $NaHCO_3$

So, acid consumed: $x + y$

Given $17.5\text{ mL}$ of $N/10 = 0.1N$ $HCl$ is used.

Since $HCl$ is monoprotic, $\text{moles of } HCl = 0.1 \times 17.5 \times 10^{-3} = 1.75 \times 10^{-3}$

Therefore, $x + y = 1.75 \times 10^{-3}$

Reaction from phenolphthalein to methyl orange end point:

Now only $NaHCO_3$ formed from $Na_2CO_3$ is neutralized: $NaHCO_3 + HCl \to NaCl + H_2O + CO_2$

Hence acid consumed in second stage equals moles of $Na_2CO_3 = y$.

Given additional $1.5\text{ mL}$ of $0.1N$ $HCl$: $y = 0.1 \times 1.5 \times 10^{-3} = 1.5 \times 10^{-4}$

**Mass of $Na_2CO_3$: **

Molar mass of $Na_2CO_3 = 106$ g/mol.

$m = y \times 106 = 1.5 \times 10^{-4} \times 106 = 0.0159\text{ g}$

Weight percentage in 0.4 g mixture:

$\%\,Na_2CO_3 = \frac{0.0159}{0.4} \times 100 = 3.975\%$

Rounded to nearest integer: $4\%$

Final Answer:

The weight percentage of $Na_2CO_3$ is: $\boxed{4}$