You have equality, yes.

The same proof [I gave here][1] applies equally well to this situation.

Here's a generalization of both (also implied by that proof).

> Suppose $\mathcal{H}$ is a hypergraph on $[n]$ where each edge has size at least $r$, any $r$ edges share at most $1$ point, and each point is contained in at least $r$ edges.  Then if $m$ is the number of edges in $\mathcal{H}$, we need $n \leq {m \choose r}$ and this is tight.

(The modification of the proof is to consider $r$-element subsets of $V_i$.)

  [1]: http://mathoverflow.net/questions/255230/minimal-number-of-edges-for-complete-linear-hypergraphs