Integrality of complex infinite series Let $(a_n)$ be a (double-sided) sequence of complex numbers satisfying
$$\sum_{\mathbb{Z}}\vert n\vert\,\vert a_n\vert^2<\infty, \tag1$$
$$\sum_{\mathbb{Z}}a_n\bar{a}_{n+k}=\delta_0(k), \qquad \forall k\in\mathbb{Z}. \tag2$$
EDIT (revealing previously withheld information). 
A theorem of Boutet de Monvel and Gabber states that under these conditions, $\sum_{\mathbb Z}n|a_n|^2$ converges to an integer. This proof is high-tech and goes through an analysis of circle-valued functions of the circle to the circle that belong to the Sobolev space $H^{1/2}(\mathbb S^1)$. (The $a_n$ are the Fourier coefficients of the function and the integer is the "degree of the function" - a concept generalizing the classical "winding number" of a function around the unit circle).

QUESTION. 
  Can you give a proof based on basic complex analysis?

REMARK. Condition (1) is an alternative Sobolev space qualification for $f\in H^{1/2}(\mathbb{S}^1)$ in terms of the Fourier coefficients $a_n$ of $f\in L^2(\mathbb{S}^1)$. Condition (2) ensures that $f$ is circle-valued. The quantity $\sum_{\mathbb Z}n|a_n|^2$ is $(1/2\pi i)\int \bar f(z)\,f'(z)\,dz=(1/2\pi i)\int f'(z)/f(z)\,dz$, which is the winding number in the case that $f$ is differentiable.
NOTATION. Here $\mathbb{S}^1$ is the unit circle, $\bar{a}$ is complex conjugation and $\delta_0(k)$ is the Dirac-delta function $\delta_0(0)=1$ and $\delta_0(k)=0$ otherwise.
 A: As correctly pointed by Anthony in the comments below, the argument I attempted here originally did not hold water.
These issues apparently have been studied quite extensively recently. See for example J. Bourgain, One cannot hear the winding number. In particular, in the introduction to this paper, Bourgain mentions that the answer to your question is yes; the result is attributed to Boutet de Monvel and Gabber.
A: This is probably true.  Here's a long way to try to prove a special case that was already completely proven in the comments to the OP.

Let's wonder what could be said if $a_{j} = 0$ for $j < 0$.  Consider the formal power series $f(z) = \sum_{n\geq 0} a_n z^n$.  Since $|a_n|$ is summable, $f(z)$ is analytic in $|z| < 1$, and in fact it can be extended to a continuous function on $|z|\leq 1$.  Consider $f(u) \overline{f(v)}$, which has the Taylor series
$$
f(u) \overline{f(v)} = \left(\sum_{n \geq 0} a_n u^n \right) \overline{\left(\sum_{m \geq 0} a_m v^m \right) } = \sum_{n,m} a_n u^{n} \overline{a_m v^m} = \sum_{k \in \mathbb{Z}} \sum_{n\geq 0} a_n u^{n} \overline{a_{n+k} v^{n+k}},
$$
where we interpret $a_{j} = 0$ if $j < 0$.  This in turn is equal to 
$$
f(u) \overline{f(v)} = \sum_{k \in \mathbb{Z}} \overline{v}^k \sum_{n\geq 0} a_n \overline{a_{n+k}} (u\overline{v})^{n}.
$$
Specializing $u=v$ for $|u| = 1$ (or taking limits as $|u| \to 1$) shows
$$
f(u) \overline{f(u)} = \sum_{k \in \mathbb{Z}} \overline{u}^k \sum_{n\geq 0} a_n \overline{a_{n+k}} = 1.
$$
Thus, we have that $|f(u)| = 1$ for all $|u|=1$ (or rather, view this as a statement about limits).  This implies that $f(u)$ is a finite Blaschke product, meaning it is of the form
$$
f(u) = c \prod_{i=1} ^{N} \left( \frac{u - r_i}{1-\overline{r_i}u}\right)^{m_i},
$$
where $m_i$ are non-negative integers, $c$ is a constant of modulus $1$ and the $r_i$ are constants of modulus less than $1$.

From here, I suspect it might be possible to finish the proof by pushing pretty hard.  Let's see what happens for the first non-trivial case, which is $f(z) = \frac{z-r}{1-rz}$, where $|r|<1$ is a real number parameter.  This has Taylor series
$$
f(z) = -r + \sum_{n=0} ^{\infty} z^{n+1} r^n (1-r^2).
$$
So our corresponding sequence will be $$a_{j} = \begin{cases}-r, \qquad &\text{if $j=0$}\\ r^{j-1} (1-r^2), \qquad &\text{if $j> 0$}\\ 0, \qquad &\text{if $j<0$} \end{cases}$$
This sequence satisfies $\sum_{n} n |a_n|^2 < \infty$ (clearly since $|r| < 1$), and $\sum_{n} a_{n}^2 = r^2 + (1-r^2)^2 \sum_{n=0}^{\infty}r^{2n} = 1,$ and for $k > 0$ (which is all we need to check by symmetry) we have
$$
\sum_{n} a_{n} \overline{a_{n+k}} = -r \cdot r^{k-1}(1-r^2) + \sum_{n\geq 0} r^{n}(1-r^2)r^{n+k}(1-r^2) = 0.$$
And we have
$$
\sum_{n} n |a_n|^2 = \sum_{n \geq 1} n (1-r^2)^2 r^{2(n-1)} = 1,
$$
as we wanted.

The fact that it works out for this first case is a very positive sign, but not yet a proof.
