= Solution
For a closed oriented $d$-dimensional <manifold> $M$ and a commutative coefficient ring $R$, <Poincare duality> says that <cap product> with the <fundamental class> is an isomorphism
$$
\boxed{-\frown[M]:H^q(M;R)\overset{\cong}{\longrightarrow}H_{d-q}(M;R).}
$$
Over a field it equivalently gives a nondegenerate <Poincare duality pairing> $(a,b)\mapsto\langle a\smile b,[M]\rangle$ between complementary cohomological degrees.
For the six-manifold take rational coefficients and write $b_j=\dim_{\mathbb Q}H^j(X;\mathbb Q)$. <Poincare duality> gives $b_j=b_{6-j}$, so the <Euler characteristic> is
$$
\chi(X)=2(b_0-b_1+b_2)-b_3.
$$
The middle-degree <Poincare duality pairing> on $H^3(X;\mathbb Q)$ is skew-symmetric by <graded commutativity of the cup product>, since $(-1)^{3\cdot3}=-1$. It is a nondegenerate <alternating bilinear form>, so its dimension $b_3$ is even. For example, a nonsingular skew-symmetric matrix of odd size would have $\det A=\det(-A)=-\det A$, impossible over $\mathbb Q$. Therefore \b[$\boxed{\chi(X)\in2\mathbb Z}$.]
To realize every even integer, use $\chi(S^6)=2$, $\chi(\mathbb{CP}^3)=4$ and $\chi(S^3\times S^3)=0$. All three are closed connected orientable six-manifolds. For the <connected sum of oriented manifolds>, removing a ball from each summand and gluing their boundary spheres gives
$$
\chi(M\#N)=\chi(M)+\chi(N)-2
$$
in dimension six: each removed open ball decreases the Euler characteristic by one, while the gluing sphere $S^5$ has Euler characteristic zero. If $k=2\ell$, put
$$
r=\max(\ell-1,0),\qquad s=\max(1-\ell,0),\qquad X_k=S^6\#\bigl(\#^r\mathbb{CP}^3\bigr)\#\bigl(\#^s(S^3\times S^3)\bigr),
$$
omitting zero copies. The <Euler characteristic> is $2+2r-2s=k$. This supplies \b[a closed connected orientable example for every $k\in2\mathbb Z$].
Without orientability, evenness need not hold. The <Real projective space> $\mathbb{RP}^6$ is a closed six-manifold with one cell in each dimension $0,\ldots,6$, so
$$
\boxed{\chi(\mathbb{RP}^6)=1.}
$$
It is nonorientable because the antipodal deck transformation on $S^6$ has degree $(-1)^7=-1$ and reverses orientation. Thus it gives the required counterexample.
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