Grassmann paradox weirdness
A supernumber $z=z_B+z_S$ consists of a body $z_B$ (which always belongs to $\mathbb{C}$) and a soul $z_S$ (which only belongs to $\mathbb{C}$ if it is zero), cf. Refs. 1 and 2.
A supernumber can carry definite Grassmann parity. In that case, it is either $$\text{Grassmann-even/bosonic/a $c$-number},$$ or $$\text{Grassmann-odd/fermionic/an $a$-number},$$ cf. Refs. 1 and 2.$^{\dagger}$ The letters $c$ and $a$ stand for commutative and anticommutative, respectively.
One can define complex conjugation of supernumbers, and one can impose a reality condition on a supernumber, cf. Refs. 1-4. Hence one can talk about complex, real and imaginary supernumbers. Note that that does not mean that supernumbers belong to the set of ordinary complex numbers $\mathbb{C}$. E.g. a real Grassmann-even supernumber can still contain a non-zero soul.
An observable/measurable quantity can only consist of ordinary numbers (belonging to $\mathbb{C}$). It does not make sense to measure a soul-valued output in an actual physical experiment. A soul is an indeterminate/variable, i.e. a placeholder, except it cannot be replaced by a number to give it a value. A value can only be achieved by integrating it out!
In detail, a supernumber (that appears in a physics theory) is eventually (Berezin) integrated over the Grassmann-odd (fermionic) variables, say $\theta_1$, $\theta_2$, $\ldots$, $\theta_N$, and the coefficient of the fermionic top monomial $\theta_1\theta_2\cdots\theta_N$ is extracted to produce an ordinary number (in $\mathbb{C}$), which in principle can be measured.
E.g. the Grassmann-odd (fermionic) variables $\psi(x,t)$ in the QED Lagrangian should eventually be integrated over in the path integral.
References:
planetmath.org/supernumber.
Bryce DeWitt, Supermanifolds, Cambridge Univ. Press, 1992.
Pierre Deligne and John W. Morgan, Notes on Supersymmetry (following Joseph Bernstein). In Quantum Fields and Strings: A Course for Mathematicians, Vol. 1, American Mathematical Society (1999) 41–97.
V.S. Varadarajan, Supersymmetry for Mathematicians: An Introduction, Courant Lecture Notes 11, 2004.
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$^{\dagger}$ In this answer, the words bosonic (fermionic) will mean Grassmann-even (Grassmann-odd), respectively.
The Lagrangian may be proved to be real but the individual factors in its terms, such as $\psi$, are neither real nor complex. They're anticommuting. There are no "particular" elements of this set of anticommuting numbers that one could "enumerate" (except for zero) and they can't appear as final predictions for observable quantities but it still makes a perfect sense to do algebra with them. A product of an even number of anticommuting variables is commuting which means that it may take particular values that may be measured and compared with theoretical predictions.
I think that I am not the only one who doesn't really understand what you're asking about but there is a chance that the answer is either in the previous paragraph or the text below:
http://motls.blogspot.com/2011/11/celebrating-grassmann-numbers.html?m=1
Let us sort out some terminology issues first. If the fermionic fields in your Lagrangian are Grassmanian, that means that the Lagrangian is classical, i.e. second quantization has not been performed yet. You can write a classical Lagrangian using c-number fermionic fields, but, as far as I understand, it is generally recognized now that one should use the classical Lagrangian with Grassmanian fermionic fields.
I also ran into the issue that you describe some time ago. I may be mistaken, but my conclusion was that indeed, the Lagrangian is not real, for the reasons that you give in your question. On the other hand, it is not obvious why this is necessarily bad.
EDIT: Maybe, to avoid ambiguity, I should have written that the Lagrangian is not real-valued