The answer is that $1 \leq a \leq 11$ Name the squares from the top left to the bottom right corner from $1$ to $9$, and let the value that the $i^{th}$ square possesses be $a_i$. Color the square in the following way: [asy][asy] import olympiad; pair A=origin, B=(1,0), C=(1,1), D=(0,1); path p =A--B--C--D--cycle; for(int i =0;i <= 2;++i){ for(int j = 0;j <= 2;++j){ if(j % 2 == 0){ filldraw(shift(i,j)*p,red); }else{ draw(shift(i,j)*p); } } } [/asy][/asy] Now let's take a look at $a_1+a_2+a_3+a_7+a_8+a_9-a_4-a_5-a_6$ Notice that when we pick any sub-square of size $2$, then we must have that $a_i+a_{i+1}-a_{i+3}-a_{i+4}$, where $i < 6$, must stay the same. Thus we have that $a_1+a_2+a_3+a_7+a_8+a_9-a_4-a_5-a_6$ must be invariant. Then we must have that $3a=a_1+a_2+a_3+a_4+\dots+a_7+a_8+a_9-2(a_4+a_5+a_6)=45-2(a_4+a_5+a_6)$ Thus we must have that $a_4+a_5+a_6$ must be divisible by $3$. This gives that $45-2.6 \geq 3a \geq 45-2.21$ This implies that $33 \geq 3a \geq 3$, which gives us that $11 \geq a \geq 1$.

@above just an obvious thing I wanted to point out is that $a$ is $ 1 \pmod{4}$....