Let wave function $\Psi$ be defined on domain $D \in \mathbb{R}^n$. The Neumann condition $\frac{\partial \Psi} {\partial {\bf n}} = 0$ on the boundary $\partial D$ has a simple interpretation in terms of the probability current of $\Psi$. For $\Delta \Psi = i \partial\Psi/\partial t$ (although it's usually taken as $i \partial\Psi/\partial t = - \Delta \Psi$), the probability current at an arbitrary point ${\bf x} \in \mathbb{R}^n$ is
$$
{\bf j}({\bf x}) = i [ \Psi^*({\bf x}){\bf \nabla}\Psi({\bf x}) - \Psi({\bf x}){\bf \nabla}\Psi^*({\bf x}) ]
$$
and the normal current on $\partial D$ reads
$$
{\bf n} \cdot {\bf j} = i\; [ \Psi^* \frac{\partial \Psi}{\partial {\bf n}} - \Psi \frac{\partial \Psi^*}{\partial {\bf n}} ]
$$
(has the wrong sign, I know, but I accounted for OP's form of the Sch.eq. as $\Delta \Psi = i \partial\Psi/\partial t$).
Setting $\frac{\partial \Psi} {\partial {\bf n}} = 0$ amounts to ${\bf n} \cdot {\bf j} = 0$ everywhere on $\partial D$, thus confining the corresponding system within $D$ without an infinite potential well, as under Dirichlet conditions ($\Psi = 0$ on $\partial D$). This is the case of perfect reflection on $\partial D$.
There is a mention of this in Section 5.2 of Visual Quantum Mechanics:
Selected Topics with Computer-Generated Animations of Quantum-Mechanical Phenomena, by Bernd Thaller (Springer, 2000); Google Books link.
As for applications, one answer to another post, Can we impose a boundary condition on the derivative of the wavefunction through the physical assumptions?, pointed to the use of Neumann conditions in R-matrix scattering theory.
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Clarification following @arivero's observation on conditions necessary to trap the system within domain D:
We can say that the system described by $\Psi$ is trapped in domain D if the total probability to locate it in D, $P_D$, is conserved in time: $dP_D/dt = 0$. In this case, if the system is initially located within D, such that $\Psi({\bf x},t=0) = 0$ for all ${\bf x} \notin D$ and $P_D(t=0) = 1$, then it will remain in D at all $t > 0$, since $P_D(t) = P_D(t=0) = 1$. If initially $P_D(t=0) < 1$ (the system has a nonzero probability to be located outside D), then we still have $P_D(t) = P_D(t=0) < 1$.
Conservation of $P_D$ is equivalent to a condition of null total probability current through the boundary $\partial D$. Note that it is not necessary to require null probability current at every point of $\partial D$, but only null total probability current through $\partial D$.
The difference can be understood in terms of path amplitudes (path-integral representation). In the former case, the amplitude $\Psi({\bf x}_1 t_1, {\bf x}_2 t_2; {\bf x} t)$ that the system "goes" from point ${\bf x}_1 \in D$ at time $t_1$ to point ${\bf x}_2 \notin D$ at time $t_2 > t_1$ while passing through point ${\bf x} \in \partial D$ at some time $t$, $t_1 < t < t_2$, is nonzero $\forall {\bf x} \in \partial D$: $\Psi({\bf x_1} t_1, {\bf x_2} t_2; {\bf x} t)≠0$. If however we demand null probability current at every point of $\partial D$, then $\Psi({\bf x}_1 t_1, {\bf x}_2 t_2; {\bf x} t)=0$, $\forall {\bf x} \in \partial D$.
In other words, null total probability current on $\partial D$ enforces weak trapping in the sense that overall $P_D(t) = $ const. and "cross-over events" across the boundary balance out. Null local probability current at every point of $\partial D$, ${\bf n} \cdot {\bf j} = 0$, corresponds to strong trapping in the sense that the system "does not cross" at any point ${\bf x} \in \partial D$. Imposing the strong trapping condition is equivalent to requiring that the weak trapping condition be satisfied by any wave function $\Psi$, as opposed to one selected $\Psi$. In this case the system is essentially confined within D at all times. Incidentally, the strong trapping condition follows from the requirement that the restriction of the system Hamiltonian on domain D remain self-adjoint.
Derivation of probability current conditions:
The free Schroedinger equation for $\Psi$, $i\partial\Psi/\partial t = \Delta\Psi$ as above (OP's choice of sign), implies local conservation of the probability density $\rho({\bf x,t}) = \Psi({\bf x},t)\Psi^*({\bf x},t)$:
$$
\frac{\partial \rho({\bf x},t)}{\partial t} + {\bf \nabla}\cdot {\bf j}({\bf x},t) = 0
$$
Integrating this over domain D yields
$$
\int_D dV\;\frac{\partial {\rho({\bf x},t)}}{\partial t} + \oint_{\partial D}dS\;{{\bf n}\cdot{\bf j}} = \frac{d}{dt}\int_DdV\;{\rho({\bf x},t)} + \oint_{\partial D}dS\;{{\bf n}\cdot{\bf j}} = 0
$$
which after denoting $P_D = \int_DdV\;{\rho({\bf x},t)}$ becomes
$$
\frac{dP_D}{dt} + \oint_{\partial D}dS\;{{\bf n}\cdot{\bf j}} = 0
$$
Imposing $dP_D/dt = 0$ necessarily means $\oint_{\partial D}dS\;{{\bf n}\cdot{\bf j}} = 0$. Note that $\oint_{\partial D} dS\;{{\bf n}\cdot{\bf j}} = 0$ does not require ${{\bf n}\cdot{\bf j}} = 0$ at every point on $\partial D$, whereas ${{\bf n}\cdot{\bf j}} = 0$ does imply $\oint_{\partial D} dS\;{{\bf n}\cdot{\bf j}} = 0$ and $dP_D/dt = 0$.
Self-adjoint restriction of the free Hamiltonian on domain D:
A self-adjoint restriction of $H\Psi = \Delta \Psi$ on D requires that $
\int_D dV\;\Phi^* (\Delta\Psi) = \int_D dV\;(\Delta\Phi^*) \Psi$ or $
\int_D dV\;[\Phi^* (\Delta\Psi) - (\Delta\Phi)^* \Psi] = 0$.
Use $\Phi^* (\Delta\Psi) = {\bf \nabla}\cdot(\Phi^* {\bf \nabla}\Psi) - {\bf \nabla}\Phi^* \cdot {\bf \nabla}\Psi$ and Green's theorem to obtain
$$
\int_D dV\;[\Phi^* (\Delta\Psi) - (\Delta\Phi)^* \Psi] = \oint_{\partial D} dS\;[\Phi^*\frac{d\Psi}{d{\bf n}} - \Psi\frac{d\Phi^*}{d{\bf n}}] = 0
$$
If the last condition above is to be satisfied by arbitrary $\Phi$, $\Psi$, it must hold locally:
$$
\Phi^* \frac{d \Psi}{d{\bf n}} - \Psi \frac{d \Phi^*}{d{\bf n}} = 0
$$
This means $\frac{1}{\Psi}\frac{d\Psi}{d{\bf n}} = \frac{1}{\Phi^*}\frac{d\Phi^*}{d{\bf n}} = a({\bf x}), \forall {\bf x} \in \partial D$. The case $\Phi = \Psi$ shows that $a({\bf x}) = a^*({\bf x})$. Therefore the restriction of $H$ on $D$ is self-adjoint if and only if wave functions in its support satisfy a boundary condition
$$
\frac{d\Psi}{d{\bf n}} = a({\bf x})\Psi
$$
for some given real-valued function $a({\bf x})$. In particular, this means every wave function $\Psi$ also satisfies the strong trapping condition ${\bf n} \cdot {\bf j} = 0$.
Finally note that the strong trapping condition means $\Psi^*\frac{d\Psi}{d{\bf n}} - \Psi\frac{d\Psi^*}{d{\bf n}} = 0$, but does not imply that $\Phi^*\frac{d\Psi}{d{\bf n}} - \Psi\frac{d\Phi^*}{d{\bf n}} = 0$ for arbitrary $\Psi$, $\Phi$. If we consider the latter expression as the matrix element of a "local current operator" ${\bf n} \cdot {\bf \hat j}$, then the strong trapping condition requires that diagonal elements of ${\bf n} \cdot {\bf \hat j}$ are 0, whereas self-adjointness requires that the entire space of wave functions is in the kernel of each ${\bf n} \cdot {\bf \hat j}$.
After making edits taking Valter Moretti's corrections into account I now feel quite confident in this answer.
I have followed your calculations and they seem correct. But you certainly can't "drop the summation" as you mention in the comments. The sum reduces (with $n = 2k$) as
$$ \left<E\right> = \sum_{k=1}^{\infty} E_n |c_n|^2
= \sum_{k=1}^{\infty} \frac{n^2 \pi^2 \hbar^2}{2mL^2} \frac{24}{n^2 \pi^2}
= \sum_{k=1}^{\infty} \frac{12 \hbar^2}{mL^2}
= \frac{12 \hbar^2}{mL^2} \sum_{k=1}^{\infty} 1 = \infty. $$
The fact is that the expectation value of the energy is actually infinite.
Such a situation might seem bizarre, but as professor Moretti pointed out it is actually not an impossible situation; you will never measure an infinite value of the energy. The probability $|c_n|^2$ of measuring the energy $E_n$ still goes to zero as $n$ goes to infinity. An infinite expectation value simply means that if you take many measurements and average them, the average will increase without bound. This does not break any particular physical principles.
Actually, as detailed in the answers to the question that Qmechanic referenced, infinite energy expectation values are typical of discontinuous wavefunctions such as the one in your initial condition. In fact, the coefficients $c_n$ that you have derived show that $\psi$ is discontinuous, because it ts a theorem of Fourier analysis that if $\psi$ is continuous, then there is a constant $K$ such that $|c_n| \leq K/n^2$. In this case, $c_n$ goes to zero as $1/n$, not $1/n^2$.
Best Answer
The domain of the problem is the entire real line, not $[0,L]$. Otherwise the potential would not be specified for $x>L, x<0$. Thus, calculate the total energy of any wavefunction that does not vanish at $x = 0,L$, you will find it to be infinite. Therefore for all eigenstates that do not have infinite energy, i.e. the entire spectrum, the wavefunction vanishes at the boundaries.