Obstruction to rationality of del Pezzo surfaces of degree 4 Let $X$ be a del Pezzo surface over a number field $k$. (A del Pezzo surface over $k$ is a smooth, projective, geometrically connected surface whose anti-canonical class $K_X$ is ample.) Let $d := K_X^2$ be the degree of $X$. It is well-known that $d$ satisfies $1 \leq d \leq 9$.
If $d \geq 5$, and $X(k) \neq \emptyset$, then $X$ is birationally equivalent to $\mathbb{P}^2_k$. Moreover, if $d=5$ or $d=7$, the condition on $X(k)$ is automatically satisfied. (This is Theorem 9.4.7 in Bjorn Poonen's Rational Points on Varieties.) In other words, for del Pezzo surfaces of degree $5$ and higher, the only obstruction to $k$-rationality is the possible lack of $k$-rational points.
My question concerns the "first" (i.e. highest-degree) non-trivial case, as far as rationality is concerned, namely the case where $d=4$. In this case, there are examples of $X$ with $X(k)\neq \emptyset$ but where $X$ is non-rational. An obstruction to rationality is given by the (non-trivial part of the) Brauer group $\operatorname{Br}(X)$ of $X$: that is, if $\operatorname{Br}(X)/\operatorname{Br}(k) \neq 0$, then $X$ is not $k$-rational. (To be sure, such $X$ exist, even among those $X$ that have rational points. This is where the $d=4$ case differs from the higher degree cases.) Indeed, this simply follows from the fact that $\operatorname{Br}$ is a birational invariant of smooth, projective, geometrically connected varieties, and from the fact that $\operatorname{Br}(\mathbb{P}^2_k) = \operatorname{Br}(k)$. My question is:

Is the converse true? That is, if $X$ is a del Pezzo surface of degree $4$ over a number field $k$ with $X(k)\neq\emptyset$, and $\operatorname{Br}(X) = \operatorname{Br}(k)$, does it follow that $X$ is a $k$-rational surface?

 A: Interesting question.
But, alas, the answer is no.
The issue is that you have missed an extra non-rationality criterion. Namely, it is possible that such a surface $X$ has $\mathrm{Br}(X) = \mathrm{Br}(k)$, yet we have $\mathrm{Br}(X_K) \neq \mathrm{Br}(K)$ for some finite extension $K/k$.
Problems of this type for cubic surfaces over finite fields are considered in the PhD thesis of Shuijing Li:
http://arxiv.org/abs/0904.3555
Though she only works over finite fields, the analysis for surfaces over number fields with a cyclic splitting field is more-or-less identical.
Here is how the counter-example goes. Take your DP4 $X$ and blow it up in a rational point not on a line to obtain a cubic surface $S$. We then choose this surface so that it has conjugacy class $C_7$ in the notation of Manin's table (p176 of his book); such a surface exists over $\mathbb{Q}$ by the recent resolution of the inverse Galois problem for cubic surfaces by Elsenhans and Jahnel. Its splitting field is cyclic of order $6$.
You see from the table that the Brauer group of $S$ is constant. However this surface is non-rational. This is because the Brauer group (modulo constants) becomes $(\mathbb{Z}/2\mathbb{Z})^2$ after a cubic extension, as it becomes the class $C_3$ from Manin's table.
I don't know whether this more general "extra non-rationality criterion" I mention is sufficient over number fields. That it is sufficient over finite fields is proven in Theorem 5.3.7 in Li's thesis.
To approach this general problem over number fields, I guess one could just enumerate all conjugacy classes of subgroups of the appropriate Weyl group in magma, and study which Brauer groups arise as well as the configurations of lines, and use the non-rationality criterion given in Theorem 5.3.1 in Li.
A: I will supplement Dan's nice answer by claiming that the answer is almost always no.  Specifically, almost every degree 4 del Pezzo surface over the rational numbers with a rational point is non-rational but has trivial Brauer group.
To support this, I wrote a little Magma program that picked two random quadrics in $\mathbb{P}^4$ containing the point $(1:0:0:0:0:0)$ and with integer coefficients in $[-5,5]$.  Let $X$ be the intersection of these two quadrics.  I computed the Fano scheme of lines on $X$.  If the Fano scheme is irreducible over $\mathbb{Q}$, that is, the lines are all Galois-conjugate, then a calculation shows that the Brauer group is trivial and the surface is not rational.  I ran the program 100 times and, apart from 1 time when $X$ was singular, this was always the case.  I'm sure it would be straightforward to show that the Fano scheme is always irreducible outside a thin set of the space parametrising such pairs of quadrics. 
