Answer: $\frac{(n+1)(n+2)}{2} - 3$.

Denote by an $(n, k)$-trapezoidal array the figure obtained by removing from an $n$-th order triangular array an $n-k$-th order triangular array sharing a common vertex with it. Let $A_{00}, A_{01}, A_{02}, \dots, A_{0n}$ denote in order the oranges on one base edge of the $n$-th order triangular array, and recursively define the orange labeled $A_{ij}$ so that $A_{ij}, A_{(i-1)j}, A_{(i-1)(j+1)}$ form a small equilateral triangle ($A_{ij} \neq A_{(i-2)(j+1)}$); we say that the orange $A_{ij}$ is the $j$-th orange in the $i$-th row.

(1) First we prove the upper bound. It suffices to prove the following:
Among $A_{00}, A_{01}, A_{02}, \dots, A_{0n}$, there must be at least one orange that is never selected.
This shows that the least upper bound must be strictly less than $\frac{(n+1)(n+2)}{2}$, and since the least upper bound must be a multiple of $3$, the result follows.

Proof of the Lemma: Suppose otherwise, i.e., assume that all these $n+1$ oranges can be removed.

* Let $k$ be the smallest integer such that $(A_{0(k+1)}, A_{0k}, A_{1(k-1)})$ are simultaneously selected; if no such integer exists, let $k = n + 1$;
* Let $m$ be the largest integer less than $k$ such that $(A_{0(m-1)}, A_{0m}, A_{1m})$ are simultaneously selected; if no such integer exists, let $m = -1$;
* Since $A_{1(k-1)}$ and $A_{1m}$ must be selected in different rounds, we must have $k - 1 \ge m + 1$. This guarantees the existence of $m < i < k$, and note that these $A_{0i}$ are not selected in either of the two selections mentioned above.

* Also note that, by the minimality of $k$ and the maximality of $m$, the selection that includes $A_{0i}$ cannot simultaneously include another $A_{0j}$, and must include at least one $A_{1j}$. But note that there are a total of $k-m-1$ such $A_{0i}$, while between $A_{1(k-1)}$ and $A_{1m}$ there are only $(k-1)-m-1$ possible $A_{1j}$'s to choose, so this is impossible. Contradiction!

(2) Next we construct a selection method that achieves the upper bound. We proceed by induction:
* $n=1$: no orange needs to be removed
* $n=2$: remove $(A_{20}, A_{10}, A_{11})$
* $n=3$: remove $(A_{30}, A_{20}, A_{11})$; $(A_{21}, A_{12}, A_{02})$
* Suppose the case $n=k-3$ holds; consider $n=k$: first apply the induction hypothesis to the first $k-3$ rows. Consider the parity of $k$:
* If $k=2m+1$, we may select:
(A$_{30}$, A$_{21}$, A$_{22}$); (A$_{20}$, A$_{11}$, A$_{12}$); (A$_{10}$, A$_{01}$, A$_{02}$); and
(A$_{0(2i+1)}$, A$_{0(2i+2)}$, A$_{1(2i+2)}$); (A$_{1(2i+1)}$, A$_{2(2i+1)}$, A$_{2(2i+2)}$), for $1 \le i \le m-2$, and
(A$_{3(k-3)}$, A$_{2(k-2)}$, A$_{1(k-2)}$); (A$_{0(k-2)}$, A$_{0(k-1)}$, A$_{1(k-1)}$)
* If $k=2m$, we may select: (A$_{30}$, A$_{21}$, A$_{11}$); (A$_{20}$, A$_{10}$, A$_{01}$) and
(A$_{0(2i+1)}$, A$_{0(2i+2)}$, A$_{1(2i+2)}$); (A$_{1(2i+1)}$, A$_{2(2i+1)}$, A$_{2(2i+2)}$), for $1 \le i \le m-2$, and
(A$_{3(k-3)}$, A$_{2(k-2)}$, A$_{1(k-2)}$); (A$_{0(k-2)}$, A$_{0(k-1)}$, A$_{1(k-1)}$)

By induction, the result follows.