Do it by induction on the number of steps $n$, that splitting the segment with length $k$, the sum of the resulting numbers on the blackboard won't exceed $\frac12k^2$. For $n=1$, suppose it is splitted into $2$ segments with length $x$ and $k-x$. $x(k-x)=-(x-\frac k2)^2+\frac{k^2}4<\frac12k^2$, the induction hypothesis holds. Suppose $\forall n<m$, the induction hypothesis holds. For $n=m$, suppose it is splitted into $2$ segments with length $x$ and $k-x$, the sum of the resulting numbers on the blackboard is at most $x(k-x)+\frac12x^2+\frac12(k-x)^2=\frac12(x+k-x)^2=\frac12k^2$, the induction hypothesis holds. $\therefore$ by induction, it holds for all number of steps $n$. $\therefore$ the sum of the resulting numbers on the blackboard won't exceed $\frac121^2=\frac12$.