Why does dimensional reduction of 10,6,2 to 9,5,1 have an equal number of supersymmetry? In general, the number of supersymmetry is different for each dimension since from dimensional reduction, the number of supersymmetry is usually increased.
$i.e$ 4d N=1 $\rightarrow$ 3d N=2 $\rightarrow $ 2d N=(2,2), N=(4,0) $\rightarrow$ 1d N=4

But for the process of 2d to 1d, the number of supersymmetry is equal. 
Re-checking from the table of number of supercharges, the dimensional reduction for 10d to 9d, 6d to 5d, 2d to 1d, the number of supersymmetry is equal. 
Why do these kind of things happens?
 A: As you can find on google, or in any book of supersymmetry, the number of components of a spinor in dimension $d$ is $2^{[d/2]}$. Where $[d/2]$ denotes the integer part of $d/2$. In certain dimensions you can impose a further property: the Majorana condition on your spinor, that reduces further of a factor of $2$ the number of independent (real) components.
The number of supercharges is the number of real components of the minimal spinor representation times the number of supersymmetries. i.e. for $\mathcal{N}=8$ in $d=4$ you have $2^2 \mathcal{N} = 32$ supercharges.
Now, for example, for the $10d$ case, you have to describe the massless particles as irreps of the little group $SO(8)$ which has two spinor reps $[0,0,1,0]$ and $[0,0,0,1]$ (using Dynkin label notation, for more information try to look at this).A bi-spinor on $SO(8)$ is the analog of the Dirac spinor in $4d$.
The two spinor reps have opposite chiralities and their dimension is $2^{rank(SO(8))} = 16$. when you reduce to $9d$ the little group for massless particles is $SO(7)$ which has just one spinor rep of dimension $2^4 = 16$, and no Majorana condition can be imposed. Thus, the number of supersymmetries remains the same, since the number of supercharges must remain the same.
If something remains unclear, please, don't hesitate to ask for clarifications.
