How to upper bound $h_1(t)-h_j(t)$ from the above ODE? Define a sequence of functions $h_i(t): [0,\infty) \to R$ for $i=1,2,\dots, n$. Assume that $\sum_{i=1}^n h_i(t)^2=1$ and $|h_1(0)|\le 1$.
If it satisfies the following differential equation:
$$h_j'(t)=\sum_{i=1}^n[(c_i-c_j)h_i(t)^2]h_j(t)$$
for some constants $c_i$ for $i=1,2,\dots, n$.
 A: For $j = 1,...,n$, we have:
$$\begin{align}
&\Longleftrightarrow h'_j(t).h_j(t)=\left(\sum_{i=1}^{n}(c_i-c_j)h_i^2(t)\right)h_j^2(t)\\
&\Longleftrightarrow (h_j^2(t))'=2\left(\sum_{i=1}^{n}(c_i-c_j)h_i^2(t)\right)h_j^2(t)\\
&\Longleftrightarrow \frac{(h_j^2(t))'}{h_j^2(t)}=2\left(\sum_{i=1}^{n}c_ih_i^2(t)-c_j\sum_{i=1}^{n}h_i^2(t)\right) =2\sum_{i=1}^{n}c_ih_i^2(t)-2c_j\\
&\Longleftrightarrow (\ln h_j^2(t))' =2\sum_{i=1}^{n}c_ih_i^2(t)-2c_j\tag{1}\\
\end{align}$$
Let's denote $g(t):=\sum_{i=1}^{n}c_ih_i^2(t)$ and $G(t) := \int_0^tg(u)du$, then
$$\begin{align}
(1)&\Longleftrightarrow (\ln h_j^2(t))' =2g(t)-2c_j\\
&\Longleftrightarrow \ln h_j^2(t)-\ln h_j^2(0) =2G(t)-2c_jt\\
&\Longleftrightarrow h_j^2(t)  =\exp{\left(2G(t)-2c_jt+\ln h_j^2(0)\right)}\hspace{3em} \forall j=1,...,n\tag{2}\\
\end{align}$$
From $(2)$ and by knowning $G'(t)=g(t)= \sum_{i=1}^{n}c_ih_i^2(t)$, we have
$$\begin{align}G'(t)&=\sum_{i=1}^{n}c_ih_i^2(t)\\
&=\sum_{i=1}^{n}\left(c_i.\exp{(2G(t)-2c_it+\ln h_i^2(0))}\right)\\
&=e^{2G(t)}\left(\sum_{i=1}^{n}c_i.e^{-2c_it+\ln h_i^2(0)}\right) \\
&=e^{2G(t)}\left(\sum_{i=1}^{n}c_i.h_i^2(0).e^{-2c_it}\right) \tag{3}\\
\end{align}$$
The differential equation $(3)$ can be solve easily:
$$\begin{align}
(3)&\Longleftrightarrow   \left( e^{-2G(t)} \right)' = -2\sum_{i=1}^{n}c_i.h_i^2(0).e^{-2c_it} \\
&\Longleftrightarrow    e^{-2G(t)} - 1 =\sum_{i=1}^{n}h_i^2(0).(e^{-2c_it}-1) \\
&\Longleftrightarrow    G(t)  =-\frac{1}{2}\ln \left(1+\sum_{i=1}^{n}h_i^2(0).(e^{-2c_it}-1) \right) \\
&\Longleftrightarrow    G(t)  =-\frac{1}{2}\ln \left(\sum_{i=1}^{n}h_i^2(0).e^{-2c_it} \right) \tag{4}\\
\end{align}$$
From $(2),(4)$, we have the closed-form solution of all $h_i(t)$ for $i=1,...,n$
$$\begin{align}|h_i(t)|&=\exp{\left(G(t)-c_it+\frac{1}{2}\ln h_i^2(0)\right)}\\
&=\exp{\left(-\frac{1}{2}\ln \left(\sum_{j=1}^{n}h_j^2(0).e^{-2c_jt} \right)-c_it+\frac{1}{2}\ln h_i^2(0)\right)}\\
\color{red}{h_i(t)} &\color{red}{=\frac{h_i(0)e^{-c_it}}{\sqrt{\sum_{j=1}^{n}h_j^2(0).e^{-2c_jt}}}}\hspace{3em} \forall i=1,...,n\tag{5}\\
\end{align} $$
The result $(5)$ is stronger than to find the upper bound of $h_i(t)-h_j(t)$, as you can calculate analytically  $h_i(t)-h_j(t)$.
