A statement: The self-intersection index of lagrangian submanifold $M \subset X$ is equal to Euler characteristic $\chi(M)$. How I should oriented $X$?
Let's consider some example. The null-section $M^2$ of cotangent bundle $T^{*}\!M^2$ is lagrangian submanifold of $T^{*}\!M$. Let $(q_1,q_2)$ be local coordinate on $M$ and $(p_1,p_2)$ is natural coordnate on $T^{*}\!M$. The local forms $\text{d}\,q_1 \wedge \text{d}\,q_2 \wedge \text{d}\,p_1 \wedge \text{d}\,p_2$ define the natural orientation of
$T^{*}\!M$. As I think, for this orientation the self-intersection index $M$ in $T^{*}\!M$ is equal to $\chi(M)$.
The natural symplectic structure on $T^{*}\!M$ is $\omega^2 = \text{d}\,p_1 \wedge \text{d}\,q_1 + \text{d}\,p_2 \wedge \text{d}\,q_2$.
The symplectic form also define orientation $$ \omega^2 \wedge \omega^2 = 2\,\text{d}\,p_1 \wedge \text{d}\,q_1 \wedge \text{d}\,p_2 \wedge \text{d}\,q_2 $$
So, the orientation induced by symplectic strucuture is not natural. And I should oriented $X$ by $(-1)\omega^2 \wedge \omega^2$.