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. 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 $$9 T_{L} =\sum_{1 \leq n \leq L} 9\left\lfloor\frac{n^{3}}{9}\right\rfloor =\sum_{1 \leq n \leq L} n^{3}-\#\{1 \leq n \leq L: n \equiv 1(\bmod 3)\}-8 \#\{1 \leq n \leq L: n \equiv 2(\bmod 3)\} =\left(\frac{1}{2} L(L+1)\right)^{2}-\left\lfloor\frac{L+2}{3}\right\rfloor-8\left\lfloor\frac{L+1}{3}\right\rfloor$$ 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 $3 L+10 / 3$, and the inequality $3 L+10 / 3 \leq L^{2}+L-2$ means exactly $L^{2}-2 L-16 / 3 \geq 0$ 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 $L \in\{\mathbf{1}, \mathbf{2}\}$.