Yes, this is possible. For each prime $p$ and $c \in \{0,1, \dots, p-1\}$ let $A_{c,p}=\{c+kp \mid k \in \mathbb{Z}\}$. Clearly, the set of all $A_{c,p}$ is a clutter $\mathcal C$ with ground set $\mathbb{Z}$. If we fix $p$, then the set of $A_{c,p}$ is a matching of size $p$. Since there are infinitely many primes, $\mathcal C$ has arbitrarily large finite matchings. On the other hand, no matching of $\mathcal C$ is infinite. To see this, it suffices to prove that if $A_{c_1,p_1}$ and $A_{c_2,p_2}$ are disjoint, then $p_1=p_2$. Suppose not. By shifting both sequences by $c_1$, we may assume that $c_1=0$. Choosing $k \equiv -c_2 p_2^{-1} \pmod{p_1}$, we have $c_2+kp_2 \equiv 0 \pmod{p_1}$, and so $A_{c_1,p_1} \cap A_{c_2,p_2}$ is non-empty.

Yes, this is possible. For each prime $p$ and $c \in \{0,1, \dots, p-1\}$ let $A_{c,p}=\{c+kp \mid k \in \mathbb{Z}\}$. Clearly, the set of all $A_{c,p}$ is a clutter $\mathcal C$ with ground set $\mathbb{Z}$. If we fix $p$, then the set of $A_{c,p}$ is a matching of size $p$. Since there are infinitely many primes, $\mathcal C$ has arbitrarily large finite matchings. On the other hand, no matching of $\mathcal C$ is infinite. To see this, it suffices to prove that if $A_{c_1,p_1}$ and $A_{c_2,p_2}$ are disjoint, then $p_1=p_2$. Suppose $p_1 \neq p_2$. By shifting both sequences by $c_1$, we may assume that $c_1=0$. Choosing $k \equiv -c_2 p_2^{-1} \pmod{p_1}$, we have $c_2+kp_2 \equiv 0 \pmod{p_1}$, and so $A_{c_1,p_1} \cap A_{c_2,p_2}$ is non-empty.

Yes, this is possible. For each prime $p$ and $c \in \{0,1, \dots, p-1\}$ let $A_{c,p}=\{c+kp \mid k \in \mathbb{Z}\}$. Clearly, the set of all $A_{c,p}$ is a clutter $\mathcal C$ with ground set $\mathbb{Z}$. If we fix $p$, then the set of $A_{c,p}$ is a matching of size $p$. Since there are infinitely many primes, $\mathcal C$ has matchings of arbitrarily large size. On the other hand, every matching of $\mathcal C$ is finite. To see this, it suffices to prove that if $A_{c_1,p_1}$ and $A_{c_2,p_2}$ are disjoint, then $p_1=p_2$. Suppose not. By shifting both sequences by $c_1$, we may assume that $c_1=0$. Choosing $k \equiv -c_2 p_2^{-1} \pmod{p_1}$, we have $c_2+kp_2 \equiv 0 \pmod{p_1}$, and so $A_{c_1,p_1} \cap A_{c_2,p_2}$ is non-empty.

Yes, this is possible. For each prime $p$ and $c \in \{0,1, \dots, p-1\}$ let $A_{c,p}=\{c+kp \mid k \in \mathbb{Z}\}$. Clearly, the set of all $A_{c,p}$ is a clutter $\mathcal C$ with ground set $\mathbb{Z}$. If we fix $p$, then the set of $A_{c,p}$ is a matching of size $p$. Since there are infinitely many primes, $\mathcal C$ has arbitrarily large finite matchings. On the other hand, no matching of $\mathcal C$ is infinite. To see this, it suffices to prove that if $A_{c_1,p_1}$ and $A_{c_2,p_2}$ are disjoint, then $p_1=p_2$. Suppose not. By shifting both sequences by $c_1$, we may assume that $c_1=0$. Choosing $k \equiv -c_2 p_2^{-1} \pmod{p_1}$, we have $c_2+kp_2 \equiv 0 \pmod{p_1}$, and so $A_{c_1,p_1} \cap A_{c_2,p_2}$ is non-empty.