Guiseppe Rossi's answer is excellent; I just want to add why the mentioned UV radiation modifies the spectrum of the background radiation.
(one word took the other, and it became a rather long comment.)
The hyperfine level
Neutral hydrogen in its ground state can be in two different configurations; either the proton and the electron may have parallel spins ($\uparrow\uparrow$), or they may have antiparallel spins ($\uparrow\downarrow$). When the spins are parallel, the atom has a slightly higher energy than when they're antiparallel. The atoms "wants" to make a spin flip to the lower energy configuration$^\dagger$, and will eventually do so, but since the line is forbidden, the lifetime of the parallel state is of the order $10^7\,\mathrm{yr}$.
The relative population of the states is given by the Boltzmann distribution
$$
\begin{array}{rcl}
\frac{n_1}{n_0} & = & \frac{g_1}{g_0} \, e^{-\Delta E \, / \, k_\mathrm{B} T_S} \\
& = & 3 \, e^{-0.068\,\mathrm{K} \, / \, T_S}\label{a}\tag{1},
\end{array}
$$
where subscripts 1 and 0 denote the $\uparrow\uparrow$ and $\uparrow\downarrow$ states, respectively, $n$ is the density, $g$ is the statistical weights (with $g_0,g_1 = 1,3$), $\Delta E = 5.9\times10^{-6}\,\mathrm{eV}$ is the energy difference of the states, $k_\mathrm{B}$ is the Boltzmann constant, and $T_S$ is the spin temperature, which I think is better thought of as "a number that describes the relative populations" than an actual temperature.
Departure from equilibrium
In thermal equilibrium, the spin temperature is equal to the "real", kinetic temperature. Just after decoupling of the radiation from matter at a redshift of $z\simeq1100$, the gas and the photons share the same energy, and since $T\gg 1$, we have that $n_1/n_0 \simeq 3$. But when the first stars begin to shine, they produce massive amounts of hard UV radiation which ionizes their surrounding medium. The ionized gas quickly recombines (in the beginning, at least), with $\sim2/3$ of the recombinations resulting in the emission of a Lyman $\alpha$ photon, i.e. a photon with an energy corresponding to the energy difference between the first excited state (one of the three $2P$ states) and the ground state (the $1S$ state) of the hydrogen atom (10.2 eV).
The Ly$\alpha$ photons scatter multiple times on the neutral hydrogen. Each scattering excites an atom from $1S\rightarrow 2P$, which subsequently de-excites and emits an Ly$\alpha$ photon in another direction. But since the energy difference between the $2P$ and the $1S$ state is a million times larger than between the hyperfine states, there is equal chance of ending in the $\uparrow\uparrow$ and the $\uparrow\downarrow$ state.
That is, $n_1/n_0$ is no longer $\simeq 3$, but is driven toward $\sim 1$. This is the Wouthuysen–Field effect that Guiseppe Rossi mentions; from eq. \ref{a}, you see that this corresponds to a much smaller spin temperature, and thus the factor that Guiseppe Rossi mentions becomes negative. The full equation describing the brightness (or, equivalently, the flux received) as a function of redshift can be written (e.g. Zaldarriaga et al. 2004)
$$
T(z) = 23\,\mathrm{mK} \,
\frac{T_S - T_\mathrm{CMB}}{T_S} \,
(1+\delta)
x_\mathrm{HI}(z)
\frac{\Omega_\mathrm{b}h^2}{0.02}
\left( \frac{0.15}{\Omega_\mathrm{m}h^2} \, \frac{1+z}{10} \right)^{1/2}\label{b}\tag{2}
$$
and when the $(T_S - T_\mathrm{CMB})\,/\,T_S$ factor is negative, you will get an absorption line.
(In eq. \ref{b},
$\delta$,
$x_\mathrm{HI}(z)$,
$\Omega_\mathrm{b}$,
$\Omega_\mathrm{m}$, and
$h$,
are
the local overdensity,
the neutral fraction of hydrogen,
the baryon and matter density parameter, and
the dimensionless (reduced) Hubble constant,
respectively, but this is of less importance.)
Since the observed absorption line (Bowman et al. 2018) starts to drop around an observed frequency of $\nu_\mathrm{obs} = 65\text{–}70\,\mathrm{MHz}$, and since the rest frequency of the hyperfine line is $\nu_\mathrm{rest} = 1420\,\mathrm{MHz}$, this means that the first stars appeared around a redshift of $z = \nu_\mathrm{rest}/\nu_\mathrm{obs} - 1 \simeq 20$, corresponding to an age of the Universe of $\sim 180\,\mathrm{Myr}$ (i.e. million years — the largest absorption is reached at $z\simeq 17$, or $t\simeq 200\,\mathrm{Myr}$).
Now the big question is, according to eq. \ref{b} the dip should be of the order a few tens of mK, but is in fact roughly 0.5 K, i.e. an order of magnitude larger. One possible mechanism that could produce this effect is coupling of the gas with dark matter, something which is not usually considered possible but could happen if the dark matter particle has a very small charge (Barkana et al. 2018).
Time evolution of the 21 cm signal
The figure below (from a great review by Pritchard & Loeb 2012) shows how the 21 cm signal evolves with time. The dip discussed in this answer is the orange and red part.
$^\dagger$An analogy would be two magnets aligned parallel to each other with north in the same direction, preferring to flip around, but note Ken G's comment below; the transition doesn't necessarily involve a spin flip, and the analogy is not to be taken literally, since parallel magnets are alike, whereas parallel electrons/protons have opposite charges.
