If the constrained Hamiltonian systems have a finite number of real degree of freedom(d.o.f), and if all the constraints are regular, then it is mathematically impossible to have an odd number of second-class constraints. (The proof is very similar to the reason why a symplectic manifold or vector space must be even-dimensional.)
Perhaps OP is actually considering a constrained Hamiltonian field theory with an infinite number of degree of freedom and an infinite number of second-class constraints (typically because all the fields, say, $\phi(\vec{r},t)$ and $\pi(\vec{r},t)$ are labeled by a continuous index, namely the space time point $(\vec{r},t)$)? In that case, it does not make sense to label $\infty$ as an odd number.
Example: A typical example of a second-class constraints in 1+1 dimension field theory with canonical Poisson brackets $$\tag{1} \{\phi(x,t),\pi(y,t)\}~=~ \delta(x-y), $$ $$\tag{2} \{\phi(x,t),\phi(y,t)\}~=~0, $$ $$\tag{3} \{\pi(x,t),\pi(y,t)\}~=~0, $$ is
$$\tag{4} \chi(x,t)~:=~\pi(x,t) -\partial_x\phi(x,t). $$
Naively one may think of (4) a single second-class constraint, but it is really infinitely many second-class constraints labeled by the position $x$. Their equal-time Poisson brackets are
$$\tag{5} \Delta(x,y)~:=~\{\chi(x,t),\chi(y,t)\}~=~ 2 \delta^{\prime}(x-y) $$
with a formal$^1$ inverse
$$\tag{6} \Delta^{-1}(x,y) ~=~ \frac{1}{4}{\rm sgn}(x-y).$$
For another related example of second-class constraints in Hamiltonian field theory, see also e.g. this Phys.SE answer.
$^1$ One should impose appropriate boundary conditions at $|x| \to \infty$.