Lower bound: $m/\gcd(m,\ell-1) - 1 \le S_m(k,\ell)$ once $k \ge n-1$

This is the lower half of the closed form, proved at the level of residues.

Throughout, $m \ge 2$ is the modulus, $\ell \ge 2$ the number of summands, $k \ge 1$ the number of colour classes, $d = \gcd(m, \ell - 1)$ and $n = m/d$.

For every $m \ge 2$ and $\ell \ge 2$, if at least $n - 1$ colours are available, that is if $k \ge n - 1$, then

The hypothesis $k \ge n-1$ is exactly what the witnessing colouring costs: give each of the residues $\overline{1}, \overline{2}, \dots, \overline{n-1}$ its own class, and pad with empty classes. Each such singleton is $\ell$-sum-free because none of $1, \dots, n-1$ is a multiple of $n$.

This is the constructive half of the mission: together with the uniform upper bound it pins the value exactly, and it is what identifies $k \ge n-1$ as a sufficient supply of colours.