The answer to your question is no, that is not possible. This is a consequence of the following slightly more general result. Let $d(n)$ is the number of divisors of $n$.
Theorem Let $G$ be a finite group of order $n$. Then for any divisor $m$ of $n$, there
exist at least $d(m)$ subgroups of $G$ of order dividing $m$.
Proof We prove the statement by induction on $m$, for all finite groups $G$. It's clear for $m=1$.
For $m>1$, let $p$ be a prime dividing $m$, and let $m = p^r k$ with $\gcd(p,k)=1$.
Since $d(m) = (r+1)d(k)$, it is sufficient to prove that, for each $i$ with
$0 \le i \le r$, there are at least $d(k)$ subgroups of $G$ of order $p^i j$,
for some $j$ that divides $k$.
So fix an $i$, let $P$ be any subgroup of $G$ of order $p^i$, let $N = N_G(P)$ be the normalizer of $P$ in $G$, and let $h = \gcd(|N|,k)$.
Since $h \le k < m$ and $h$ divides $|N/P|$, by inductive hypothesis, the number $s$ of subgroups $S/P$ of $N/P$ of order dividing $h$ satisfies $s \ge d(h)$. For each of these subgroups, the inverse image $S$ of $S/P$ in $N/P$ is a subgroup of $G$ of order $p^i j$, where $|S/P| = j$ divides $h$, and so $j$ divides $k$.
Now $P$ has $|G|/|N|$ distinct conjugates $P^g$ in $G$, and the $s|G|/|N|$ subgroups $S^g$ are all distinct. But $|G|/|N| \ge k/h$, so $s|G|/|N| \ge d(h)k/h \ge d(k)$, which completes the proof.
After thinking about it again, I think we can say that a group of order $n$ with exactly $d(n)$ subgroups must be cyclic. The above proof produces at least $d(n)$ subgroups with a normal $p$-subgroup for any prime divisor $p$ of $n$. So if there were exactly $d(n)$ subgroups, then all subgroups would have all of their Sylow subgroups normal, so they would all, including $G$ itself, be nilpotent. But a non-cyclic $p$-group $P$ has more than one subgroup of index $p$, and hence has more than $d(|P|)$ subgroups, so all Sylow subgroups of $G$ are cyclic and hence so is $G$.
By the way, I first posted this proof here in 2003. I would welcome any references to any other proofs.