1. We need the oxidation numbers of the two different sulphur atoms in $Na_2S_4O_6$.

2. First, identify the anion: $Na_2S_4O_6 \rightarrow 2Na^+ + S_4O_6^{2-}$ So, we work with $S_4O_6^{2-}$.

3. In tetrathionate ion, the structure is: $^-O_3S - S - S - SO_3^-$

   Thus, there are two types of sulphur atoms:

   • the two terminal sulphur atoms attached to oxygen atoms,
   • the two inner sulphur atoms linked only to sulphur atoms.

4. Let the oxidation number of each terminal sulphur be $x$. Each terminal sulphur is bonded to three oxygens. Since oxygen is $-2$, and the terminal sulphur environment is like a sulfonate sulphur, its oxidation state is: $x = +5$

5. Let the oxidation number of each inner sulphur be $y$. These sulphur atoms are bonded only to sulphur atoms. Since bonds between identical atoms do not contribute to oxidation number change: $y = 0$

6. Check by overall oxidation number balance in $S_4O_6^{2-}$: $2(+5) + 2(0) + 6(-2) = -2$ $10 - 12 = -2$ which is correct.

7. Therefore, the two oxidation numbers are: $+5 \text{ and } 0$

8. Their difference is: $5 - 0 = 5$

Hence, the required integer answer is: $\boxed{5}$