The largest such integer $N$ is $105$. Let $p_i$ denote the $i$th prime, i.e. $p_1 = 2$, $p_2 = 3$, etc. We call a positive integer $N$ admissible if it is odd and has the property stated in the problem. Any admissible number exceeding $p_i^2$ must clearly contain the factor $p_i$ for $i \ge 2$. By Bertrand's postulate, $p_{n+1} < 2p_n$. Thus for any $n \ge 5$, we have $p_{n+1}^2 < 4p_n^2 < 8p_{n-1}p_n$. Since $p_2p_3 = 15 > 8$,
$$p_{n+1}^2 < p_2 p_3 \cdots p_n \quad (n \ge 5) \qquad (1)$$
If there were an admissible number $N$ such that $p_{n+1}^2 \ge N > p_n^2$ for some $n \ge 5$, then $N$ would be divisible by $p_2p_3 \cdots p_n$ and so $N \ge p_2p_3 \cdots p_n > p_{n+1}^2$ by (1) which gives a contradiction. Therefore no admissible number can exceed $p_5^2 = 121$.
The number $105$ can be checked to be admissible. Also since any admissible number exceeding $49$ must be divisible by $3 \times 5 \times 7 = 105$, there can be no admissible number $N$ between $105$ and $121$.