Answer already hinted at in comments...
A complete sufficient statistic does exist for this family of distributions, a Gumbel density with scale parameter unity. The set of order statistics is not minimal sufficient here. We should consider order statistics, which are trivially sufficient for any family of distributions, when we cannot find any other (non-trivial) sufficient statistic, say for the Laplace distribution and the Cauchy distribution with unknown location parameter.
Due to independence, joint density of $(X_1,X_2,\cdots,X_n)$ is
\begin{align}
f_{\theta}(x_1,x_2,\cdots,x_n)&=\prod_{i=1}^nf(x_i\mid\theta)
\\&=e^{-\sum_{i=1}^n x_i+n\theta}\exp\left(-\sum_{i=1}^n e^{-(x_i-\theta)}\right)
\\&=\exp\left(-e^{\theta}\sum_{i=1}^n e^{-x_i}+n\theta\right)e^{-\sum_{i=1}^nx_i}
\\&=g(\theta,t(\mathbf x))\,h(\mathbf x)\qquad,\text{ for all } \mathbf x=(x_1,\cdots,x_n)\in\mathbb R^n\,,\,\theta\in\mathbb R
\end{align}
where $g(\theta,t(\mathbf x))=\exp\left(-e^{\theta}\sum_{i=1}^n e^{-x_i}+n\theta\right)$ depends on $\theta$ and on $x_1,x_2,\cdots,x_n$ through $t(\mathbf x)=\sum_{i=1}^n e^{-x_i}$ and $h(\mathbf x)=e^{-\sum_{i=1}^nx_i}$ is independent of $\theta$.
So by Factorisation theorem, a sufficient statistic for $\theta$ is $$T(\mathbf X)=\sum_{i=1}^n e^{-X_i}$$
Let $Y=e^{-X}$ where $X$ has the density $f(x\mid\theta)$.
Then density of $Y$ is
\begin{align}f_Y(y)&=f(-\ln y\mid\theta)\left|\frac{dx}{dy}\right|\mathbf1_{y>0}
\\&=\frac{1}{y}e^{\ln y+\theta}\exp\left(-e^{\ln y+\theta}\right)\mathbf1_{y>0}
\\&=e^{\theta}\exp(-ye^{\theta})\mathbf1_{y>0}
\end{align}
That is, $Y\sim\text{Exp}(\lambda)$ with $\lambda=e^{\theta}(>0)$.
In other words, $e^{-X_i}\stackrel{\text{i.i.d}}\sim\text{Exp}(\lambda)$ for each $i=1,2,\cdots,n$, which implies
$$T\sim\text{Gamma}(\lambda,n)$$
To show that $T$ is indeed a complete statistic, we have to show that for any function $\psi$ of $T$, $$E_{\theta}(\psi(T))=0\implies \psi(T)=0\quad,\text{ a.e.}$$
Now,
\begin{align}E_{\theta}(\psi(T))&=0
\\\implies \int_0^\infty \psi(t)e^{-\lambda t}t^{n-1}\,dt &=0
\end{align}
It can be shown using the unicity property of Mellin transform that the last implication means $\psi(t)$ itself is identically zero. There might be an easier proof for completeness but I could not find one.
Best Answer
It turns out that after further expansion, the above becomes
$\dfrac{f_\theta(x)}{f_\theta(y)} = \exp \left\{ - \left[ \dfrac{1}{2\theta} \left( \sum x_i^2 - \sum y_i^2 \right) + \left( \sum y_i - \sum x_i \right) \right] \right\}$.
Since it is only the first term that depends on $\theta$, the M.S.S is $\sum x_i^2$.