As I understand, from the comments, the goal is to fix $d$ and then require that the additive drop between sites $s_i$ and $s_{i+1}$ be $d$.
Assume $d,n$ are fixed. Then all we need to know is the amount of traffic to the first site, call it $s_1$. Then $s_i=s_1-(i-1)d$ so we compute $$100=\sum_{i=1}^n\left(s_1-(i-1)d\right)=ns_1-\frac {(n-1)(n)}2d$$
It follows that we should take $$\boxed{s_1=\frac {100}n+\frac {n-1}2d}$$
Example: $n=3,d=20$. Then $$s_1=\frac {100}3+20=53.333\dots$$ Thus in this case your numbers should be $\\{53.333,33.333,13.333\\}$. Note that the numbers you gave do not add to $100$.