Suppose $ ad = bc$, for some $ a < b\le c < d\in S$. Let $ a = M^{2}+x$, $ b = M^{2}+y$, $ c = M^{2}+z$, $ d = M^{2}+t$. Then $ 0\le x < y\le z < t\le 2M$. $ ad = bc\Leftrightarrow M^{2}(x+t)+xt = M^{2}(y+z)+yz\Leftrightarrow M^{2}(x+t-y-z) = yz-xt$, from where $ yz-xt = kM^{2}$, for some $ k$. Because $ yz-xt\le yz < t^{2}\le4M^{2}$, we have $ 0\le k\le3$. Also $ y+z = x+t-k$. $ k = 0$ means $ xt = yz$ and $ y+z = x+t$, easily implying $ x = y$, $ z = t$, impossible. Take now $ k > 0$. Then $ 4yz\le(y+z)^{2}= (x+t-k)^{2}$, so $ 4yz = 4kM^{2}+4xt\le(x+t-k)^{2}$, or $ 4kM^{2}+2k(x+t)\le(t-x)^{2}+k^{2}\le4M^{2}+k^{2}$. This is $ 4M^{2}(k-1)+k(2x+2t-k)\le0$. But $ k\ge1$ and because $ x+2\le t$, we have $ 2x+2t\ge4 > k$, contradiction.

Other method. Suppose exist $ (a,b,c,d)\in N$ for $ ab=cd$ which $ a<c<b<b$ and Apply four number lemma we have exist $ (m,n,p,q)$ $ a=mn,b=pq,c=np,d=mq$ Because $ a<c,b>d$ imply that $ m<p,q>n$ Imply that $ p\geq m+1,q\geq n+1$ So $ pq\geq (m+1)(n+1)\geq (\sqrt{mn}+1)^2\geq (M+1)^2$ (Tradition!).