, 2002

Solution:

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 $L = 1$ or $L$ is even.