All solutions (by now) involve the Fourier transform which is quite surprising. Solution below by Yongxi Wang

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Hard problem! By DFT

I'll give the official solution, by the idea of DFT in two sequences. I'll skip few calculating steps or it is too long. For $\forall 0\le j\le s-1,$ define $x_j=\left\{\frac{jm}s\right\}\left(1-\left\{\frac{jm}s\right\}\right),y_j=\left\{\frac{jn}s\right\}\left(1-\left\{\frac{jn}s\right\}\right).$ Let $\omega =\exp\left(\frac{2\pi i}{s}\right).$ Consider the Fourier Transform of $x_j,y_j,$ for all $0\le k\le s-1,$ define $a_k=\sum_{j=0}^{s-1}x_j\omega^{kj},b_k=\sum_{j=0}^{s-1}y_j\omega^{kj}.$ Lemma 1 (Similar to Parseval). $\sum_{j=0}^{s-1}x_jy_j=\frac 1s\sum_{k=0}^{s-1}a_kb_k.$ Proof. By $x_j=x_{s-j},y_j=y_{s-j},$ we have $a_k=a_{s-k},b_k=b_{s-k}.$ Therefore \begin{align*}\sum_{k=0}^{s-1}a_kb_k&=\sum_{k=0}^{s-1}a_kb_{s-k}=\sum_{k=0}^{s-1}\sum_{j_1=0}^{s-1}x_{j_1}\omega^{kj_1}\sum_{j_2=0}^{s-1}y_{j_2}\omega^{-kj_2}\\&=\sum_{j_1=0}^{s-1}\sum_{j_2=0}^{s-1}x_{j_1}y_{j_2}\sum_{k=0}^{s-1}\omega^{k(j_1-j_2)}=s\sum_{j=0}^{s-1}x_jy_j.\Box\end{align*}Lemma 2. $\sum_{j=0}^{s-1}x_jy_j=\frac s{36}-\frac 1{18s}+\frac 1{36s^3}+\frac 1{4s^3}\sum_{k=1}^{s-1}\dfrac 1{{\sin^2\frac{km}s}\sin^2\frac{kn}s}.$ Proof.For $\forall 1\le k\le s-1,$ \begin{align*}a_k&=\sum_{j=0}^{s-1}x_j\omega^{kj}=\sum_{j=0}^{s-1}\left\{\frac{jm}s\right\}\left(1-\left\{\frac{jm}s\right\}\right)\omega^{kj}\\&=\sum_{j=0}^{s-1}\frac js\left(1-\frac js\right)\omega^{km^{-1}j}=\frac 2s\cdot\frac{\omega^{km^{-1}}}{(1-\omega^{km^{-1}})^2}=\dfrac 1{2s\sin^2\frac{km^{-1}}s}.\end{align*}Similarly $b_k=\dfrac 1{2s\sin^2\frac{kn^{-1}}s}.$ Therefore \begin{align*}\sum_{j=0}^{s-1}x_jy_j&=\frac 1s\sum_{k=0}^{s-1}a_kb_k=\frac {a_0b_0}s+\frac 1s\sum_{k=1}^{s-1}a_kb_k\\&=\frac 1s\left(\sum_{j=0}^{s-1}\frac js\left(1-\frac js\right)\right)^2+\frac 1s\cdot\frac 1{4s^2}\sum_{k=1}^{s-1}\dfrac 1{\sin^2\frac{km^{-1}}s\sin^2\frac{kn^{-1}}s}\\&=\frac s{36}-\frac 1{18s}+\frac 1{36s^3}+\frac 1{4s^3}\sum_{k=1}^{s-1}\dfrac 1{{\sin^2\frac{km}s}\sin^2\frac{kn}s}.\Box\end{align*}Back to the problem, using the result proved in 2021 CTST P9: $\sum_{k=1}^{s-1}\dfrac 1{{\sin^2\frac{km}s}}=\frac {s^2-1}3.$ So we have \begin{align*}\sum_{j=1}^{s-1}\left\{\frac{jm}s\right\}\left(1-\left\{\frac{jm}s\right\}\right)\left\{\frac{jn}s\right\}\left(1-\left\{\frac{jn}s\right\}\right)&=\frac s{36}-\frac 1{18s}+\frac 1{36s^3}+\frac 1{4s^3}\sum_{k=1}^{s-1}\dfrac 1{{\sin^2\frac{km}s}\sin^2\frac{kn}s}\\&\ge\frac s{36}-\frac 1{18s}+\frac 1{36s^3}+\frac 1{4s^3}\sum_{k=1}^{s-1}\dfrac 1{{\sin^2\frac{km}s}}\\&=\frac s{36}-\frac 1{18s}+\frac 1{36s^3}+\frac 1{4s^3}\cdot\frac{s^2-1}3>\frac s{36}.\end{align*}Also obviously it is sharp (same as above), so the maximum of $c$ is $\frac 1{36}.\Box$