HMMT February

We distinguish two cases depending on the parity of $L$. Suppose first that $L=2k-1$ is odd, where $k \geq 1$. Then $S_{L}=\sum_{1 \leq n \leq 2k-1}\left\lfloor\frac{n}{2}\right\rfloor=2 \sum_{0 \leq m<k} m=2 \cdot \frac{k(k-1)}{2}=k(k-1)$. If $k=1$, this is the square number 0. If $k>1$ then $(k-1)^{2}<k(k-1)<k^{2}$, so $k(k-1)$ is not square. Now suppose $L=2k$ is even, where $k \geq 1$. Then $S_{L}=S_{L-1}+k=k^{2}$ is always square. Hence $S_{L}$ is square exactly when $L=\mathbf{1}$ or $L$ is even.