jmc

There must be some polynomial $Q(x)$ such that $$P(x)-a=(x-1)(x-3)(x-5)(x-7)Q(x).$$Then, plugging in values of $2,4,6,8,$ we get

$$P(2)-a=(2-1)(2-3)(2-5)(2-7)Q(2)=-15Q(2)=-2a,$$$$P(4)-a=(4-1)(4-3)(4-5)(4-7)Q(4)=9Q(4)=-2a,$$$$P(6)-a=(6-1)(6-3)(6-5)(6-7)Q(6)=-15Q(6)=-2a,$$$$P(8)-a=(8-1)(8-3)(8-5)(8-7)Q(8)=105Q(8)=-2a.$$That is, $$-2a=-15Q(2)=9Q(4)=-15Q(6)=105Q(8).$$Thus, $a$ must be a multiple of $\text{lcm}(15,9,15,105)=315$.

Now we show that there exists $Q(x)$ such that $a=315.$ Inputting this value into the above equation gives us $$Q(2)=42,Q(4)=-70,Q(6)=42,Q(8)=-6.$$From $Q(2)=Q(6)=42,$ $Q(x)=R(x)(x-2)(x-6)+42$ for some $R(x).$ We can take $R(x)=-8x+60,$ so that $Q(x)$ satisfies both $Q(4)=-70$ and $Q(8)=-6.$

Therefore, our answer is $\boxed{315}.$