, 2002

Solution:
Since $T_{L}$ is square if and only if $9 T_{L}$ is square, we may consider $9 T_{L}$ instead of $T_{L}$.
It is well known that $n^{3}$ is congruent to $0, 1$, or $8$ modulo $9$ according as $n$ is congruent to $0, 1$, or $2$ modulo $3$. (Proof: $(3m + k)^{3} = 27m^{3} + 3(9m^{2})k + 3(3m)k^{2} + k^{3} \equiv k^{3} \pmod{9}$.) Therefore
$n^{3} - 9\left\lfloor n^{3} / 9 \right\rfloor$ is $0, 1$, or $8$ according as $n$ is congruent to $0, 1$, or $2$ modulo $3$. We find therefore that
$$\begin{aligned} 9 T_{L} &= \sum_{1 \leq n \leq L} 9\left\lfloor \frac{n^{3}}{9} \right\rfloor \\ &= \sum_{1 \leq n \leq L} n^{3} - \#\{1 \leq n \leq L : n \equiv 1 \pmod{3}\} - 8\#\{1 \leq n \leq L : n \equiv 2 \pmod{3}\} \\ &= \left( \frac{1}{2} L(L+1) \right)^{2} - \left\lfloor \frac{L+2}{3} \right\rfloor - 8\left\lfloor \frac{L+1}{3} \right\rfloor . \end{aligned}$$
Clearly $9 T_{L} < (L(L+1)/2)^{2}$ for $L \geq 1$. We shall prove that $9 T_{L} > (L(L+1)/2 - 1)^{2}$ for $L \geq 4$, whence $9 T_{L}$ is not square for $L \geq 4$. Because
$$(L(L+1)/2 - 1)^{2} = (L(L+1)/2)^{2} - L(L+1) + 1,$$
we need only show that
$$\left\lfloor \frac{L+2}{3} \right\rfloor + 8\left\lfloor \frac{L+1}{3} \right\rfloor \leq L^{2} + L - 2 .$$
But the left-hand side of this is bounded above by $3L + 10/3$, and the inequality $3L + 10/3 \leq L^{2} + L - 2$ means exactly $L^{2} - 2L - 16/3 \geq 0$ or $(L-1)^{2} \geq 19/3$, which is true for $L \geq 4$, as desired.
Hence $T_{L}$ is not square for $L \geq 4$. By direct computation we find $T_{1} = T_{2} = 0$ and $T_{3} = 3$, so $T_{L}$ is square only for $L \in \{\mathbf{1}, \mathbf{2}\}$.