What is the origin of the factor of $-1/4$ in the Maxwell Lagrangian? I have seen numerous 'derivations' of the Maxwell Lagrangian,
$$\mathcal{L} ~=~ -\frac{1}{4}F_{\mu \nu}F^{\mu \nu},$$
but every one has sneakily inserted a factor of $-1/4$ without explaining why. The Euler-Lagrange equations are the same no matter what constant we put in front of the contraction of the field strength tensors, so why the factor of $-1/4$?
 A: The interesting point is that with factor -1/4 $$ \mathcal L =-\frac 1 4 F_{\mu\nu}F^{\mu\nu}= \frac 1 2 (\mathbf E^2- \mathbf B^2)\;,  $$
From this we roughly have
$$\mathcal H =\frac 1 2 (\mathbf E^2+ \mathbf B^2)\;, $$
which consistence with the energy of the EM field$^{[1]}$
$$U=\frac 1 2 \int \big(  \epsilon_0 \mathbf E^2 + \frac 1 {\mu_0}\mathbf B^2 \big) d\tau \;.$$
References
[1] D.J. Griffiths, Introduction to electrodynamics Fourth Edition (See back materials)
A: Comments to the question:


*

*First it should be stressed, as OP does, that the Euler-Lagrange equations (= classical equations of motion = Maxwell's equations) are unaffected by scaling the action $S[A]$ with an overall (non-zero) constant. So classically, one may choose any overall normalization that one would like.

*As Frederic Brünner mentions a normalization of the $J^{\mu}A_{\mu}$ source term with a normalization constant $\pm N$ goes hand in hand with a $-\frac{N}{4}$ normalization of the $F_{\mu\nu}F^{\mu\nu}$ term. Here the signature of the Minkowski metric is $(\mp,\pm,\pm,\pm)$. 

*Recall that the fundamental variables of the Lagrangian formulation are the $4$-gauge potential $A_{\mu}$. Here $A_{0}$ is a non-dynamical Lagrange multiplier. The dynamical variables of the theory are $A_1$, $A_2$, and $A_3$. The $$-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}~=~\underbrace{\frac{1}{2} \sum_{i=1}^3\dot{A}_i\dot{A}_i}_{\text{kinetic term}}+\ldots$$ is just the standard $+\frac{1}{2}$ normalization of a kinetic term in field theory. In particular note that the kinetic term is positive definite in order not to break unitarity.
A: The factor is there so that once you add a source term, i.e. $J^\mu A_\mu, $ you get the correct equations of motion, namely Maxwell's equations:
$\partial_\nu F^{\mu\nu}=J^\mu.$
Furthermore, this convention produces the usual $1/2$ in front of the kinetic term of the gauge fields. 
