Answer. $1$.

First solution. By induction on $N$, we prove that the number of even matchings is one more than the number of odd matchings. For $N=1$ the statement is obvious: there is only one matching, and it is even. Now, prove the statement for $2N$ points, assuming it is true for $2(N-1)$ points. Denote the marked points $A_1, A_2, \dots, A_{2N}$ in order around the circle clockwise.

Lemma. Suppose a matching contains the chord $A_1A_i$. Then for even $i$ it intersects an even number of chords, and for odd $i$ — an odd number.
Proof. Suppose the chord $A_1A_i$ is intersected by exactly $k$ chords. Consider the points $A_2, \dots, A_{i-1}$; exactly $k$ of them are endpoints of chords intersecting $A_1A_i$ (one endpoint for each such chord). The remaining $i-2-k$ points are split into pairs of points joined by chords that do not intersect $A_1A_i$. Thus, the number $i-2-k$ is even, so $i$ and $k$ have the same parity. The lemma is proved. $\square$

Now, divide all matchings into $2N-1$ groups $\Pi_2, \dots, \Pi_{2N}$: group $\Pi_i$ contains those matchings in which point $A_1$ is joined to $A_i$. Now, remove the chord $A_1A_i$ from each matching in $\Pi_i$; we get all possible matchings on the remaining $2N-2$ points. By the induction hypothesis, among them, the number of even matchings is one more than the number of odd matchings. Moreover, if $i$ is even, then by the lemma the parity of the matching does not change when removing the chord, and if $i$ is odd, it does change. Thus, in each of the $N$ sets $\Pi_2, \dots, \Pi_{2N}$, the number of even matchings is one more than the number of odd, and in each of the $N-1$ sets $\Pi_3, \dots, \Pi_{2N-1}$, the number of odd matchings is one more than the number of even. In total, the number of even matchings exceeds the number of odd by $N-(N-1)=1$, as required.

Second solution. We give another proof of the induction step.
Let the marked points be $A_1, \dots, A_{2N}$. Consider all matchings in which $A_{2N-1}$ and $A_{2N}$ are joined by a chord. This chord does not intersect any other. Thus, removing it from each such matching, we obtain all matchings on the points $A_1, \dots, A_{2N-2}$, and the parity of each is preserved.

By the induction hypothesis, among these matchings, the number of even is one more than the number of odd.
To complete the proof, it suffices to show that among all other matchings, the numbers of even and odd are equal. Consider any such matching; suppose it contains chords $A_{2N-1}A_i$ and $A_{2N}A_k$. Now "swap" the points $A_{2N-1}$ and $A_{2N}$, i.e., replace these chords with $A_{2N}A_i$ and $A_{2N-1}A_k$. If the original chord intersected some of the others, so does the new chord. On the other hand, if the chords $A_{2N-1}A_i$ and $A_{2N}A_k$ did not intersect, the new chords will intersect, and vice versa. Thus, to each remaining even matching we associate an odd one, and vice versa; and different matchings correspond to different ones. Therefore, the numbers of even and odd matchings among the rest are equal, as required.