Let's denote an operation of sequences: From $(x_1, x_2, \cdots, x_k)$, select two equal values $x_i=x_j$, and replace to $x_i-1$ and $x_j+1$. If $x_i-1$ becomes zero, erase it. Lemma. If $\displaystyle\frac{k(k+1)}{2}<n\leq \displaystyle\frac{(k+1)(k+2)}{2}$, it is possible to make a sequence with all integers from $1$ to $k+1$ except $\displaystyle\frac{(k+1)(k+2)}{2}-n$ by some operations from $(1, 1, \cdots, 1)$($n$ times). pf) Easily shown by induciton of $n$. Since $(x+1)^2+(x-1)^2=x^2+2$, the sum of squares of the terms increases by $2$ by each operation. As the initial sum is $n$, all sums from $n$ to the last value calculated by the Lemma cannot be $S(n)$. $\therefore S(n)>\displaystyle\sum_{i=1}^{k+1}{i^2} - (\displaystyle\frac{(k+1)(k+2)}{2}-n)^2>\displaystyle\frac{(k+1)(k+2)(2k+3)}{6}-(k+1)^2$ Since $k=O(\sqrt{n})$, we can find $n_0$ and $c>0$ such that $S(n)\geq c n^{1.5}$ for $n\geq n_0$.

Consider the following algorithm: Starting at $(1, 1,\dots, 1)$, given the k-tuple $(x_1, x_2, \dots, x_k)$: $\bullet$ If there are $x_i=x_j$ we make the following change: $(x_1, x_2, \dots, x_k)\longmapsto (x_1, x_2, \dots, x_i-1, x_j+1, x_k)$, else we stop. Notice that $\sum_{m=0}^k x_m=n$, and on each step the sum $\sum_{m=0}^k x_m^2$ increases by $2$. If we stop at $(y_1, y_2,\dots, y_k)\Rightarrow n\geq \frac{k\cdot (k+1)}{2}\Rightarrow n>\frac{k^2}{2}\Rightarrow k<\sqrt{2n}$ Then, $y_1+y_2+\dots+y_k\geq 1+2+\dots +k\Rightarrow y_1^2+y_2^2+\dots+y_k^2\geq \frac{(y_1+y_2+\dots+y_k)^2}{k}\geq \frac{n^2}{\sqrt{2n}}=\frac{1}{\sqrt{2}}\cdot n^{\frac{3}{2}}$. But, because we covered all the numbers less than $\frac{1}{\sqrt{2}}\cdot n^{\frac{3}{2}}$, we are done.$_{\blacksquare}$

nice problem