Bumppp this

the answer is $850$. the construction is as follows : $$\underbrace{00\ldots 00}_{18} 1 \underbrace{0000}_{4} 1 \underbrace{0000}_{4} 1 \ldots 1 \underbrace{0000}_{4} 1$$ where there are $17$ numbers of $1$.

Proof that $850$ is optimal: Obviously, we can change all numbers to zeros and ones. We start with zeros and finish with a one. Let us have $k$ ones and $a_1,a_2,\ldots,a_k$ be zeros in between ones. Then if $a_i<4$ for $i \geq 2$, then consider $a_i+2$ consecutive numbers and get a contradiction. Thus, for $i \geq 2$ we know $a_i \geq 4$. Also $a_1+\ldots+a_k+k=99$. $|S|=ka_1+(k-1)a_2+\ldots+1a_k=k(a_1+\ldots+a_k)-(a_2+2a_3+\ldots+(k-1)a_k) \leq k(99-k)-(4+2*4+\ldots+(k-1)*4)=99k-k^2-2(k-1)k=-3k^2+101k$. It's peak is at $k=\frac{101}{6}$, which is around $16.8$. So $k=17$ is the best, thus $|S| \leq -3*17^2+101*17=17*(-51+101)=850$, which is the desired.