Answer is $49$. For $M\geq 49$, we set $a_{2k+1}\rightarrow 0$ and $a_{2k}=1$. Let's show that $M=49$ holds. Claim: If $f(a_k)=\frac{a_k}{a_{k-1}+a_k+a_{k+1}}$, then there exists $f(a_{k-1})+f(a_k)+f(a_{k+1})\leq 1$. Proof: Suppose not. For $a_{k-2}=a,a_{k-1}=b,a_k=c,a_{k+1}=d,a_{k+2}=e$, we have \[\frac{b}{a+b+c}+\frac{c}{b+c+d}+\frac{d}{c+d+e}>1\iff \frac{b}{a+b+c}+\frac{d}{c+d+e}>\frac{b+d}{b+c+d}\]\[\iff \frac{b(b+c+d)}{a+b+c}-b+\frac{d(b+c+d)}{c+d+e}-d>0\iff \frac{b(d-a)}{a+b+c}+\frac{d(b-e)}{c+d+e}>0\]Thus, there doesn't exist $a_k\geq a_{k+3}$ and $a_{k+4}\geq a_{k+1}$. Since $99$ is divisible by $3$, there exists $a_m\geq a_{m+3}$ (consider $a_3,...,a_{99})$. Then, $a_{m+1}>a_{m+4}$ which yields $a_{m+2}>a_{m+5}$. We must continue this proess but $a_3>a_6>...>a_{99}>a_3$ is impossible.$\square$ Claim: $f(a_k)+f(a_{k+1})\leq 1$. Proof: \[\frac{a_k}{a_{k-1}+a_k+a_{k+1}}+\frac{a_{k+1}}{a_k+a_{k+1}+a_{k+2}}\leq \frac{a_k}{a_k+a_{k+1}}+\frac{a_{k+1}}{a_k+a_{k+1}}=1\]As we have expected.$\square$ WLOG $f(a_1)+f(a_2)+f(a_3)\leq 1$. \[\sum{f(a_i)}\leq 1+48=49\]As desired.$\blacksquare$

I think it is a little bit hard for P5.