We shall prove this general with $p$ red boxes and $q$ blue boxes and $pq$ objects, with $p, q$ coprime. Suppose the bags have been distributed as described, with each red box containing $q$ objects and each blue box containing $q$ objects. Suppose we have a collection of $r$ red and $b$ blue boxes. We say that a bag is internal (with respect to the collection) if it is contained in a red box and a blue box in the collection. It is external if it is contained in only one of the boxes in the collection. If the collection does not have any external bags, then $rq = bp$. Since $p, q$ are coprime, we have $r = p, b = q$. This means any collection of $x$ boxes, with $x < p+q$, must have an external bag.

We shall prove that for every $x < p+q$, there is a collection of $r$ red and $b$ blue boxes with $r+b = x$ that has at least $x-1$ internal bags and at least 1 external bag. This is certainly true for $x = 2$ since every bag is contained in a red box and a blue box and that these 2 boxes have an external bag. Inductively, suppose $x < p+q-1$ and there is a collection of $x$ boxes with $x-1$ internal bags and 1 external bag. By putting the box that contains the external bag into the collection, we have a collection of $x+1$ boxes with $x$ internal bags. Since $x+1 < p+q$, the collection also has an external bag. This can continue until $x = p+q-1$. For this collection, we have $x-1$ internal and 1 external bag giving a total $x = p+q-1$ bags. Thus $n \ge p+q-1$.

Next we shall show that $n = p+q-1$ is sufficient. Apply the division algorithm, assuming that $p > q$, we have $p = kq+r$ where $0 \le r < q-1$. Suppose that we have solved the problem for $(q, r)$, i.e., we have $q+r-1$ bags containing a total of $qr$ objects and these bags can be distributed into $q$ red boxes, $R_1, \dots, R_q$ with $r$ objects in each box and into $r$ blue boxes, $B_1, \dots, B_r$, with $q$ objects in each box. Now add another $kq$ bags with $q$ objects in each bag. We have a total $kq + q + r - 1 = p + q - 1$ bags with a total of $kq^2 + qr = pq$ objects. Put each of these new bags into a new blue box, then, together with $B_1, \dots, B_r$, we have $kq + r = p$ blue boxes each with $q$ objects. Next into each of the red boxes $R_1, \dots, R_q$, put $k$ of these new bags so that each box has $kq + r = p$ objects. Thus we have a solution for $(p, q)$. Since $p, q$ are coprime, applying the division algorithm repeatedly, we eventually get to the situation where $r=1$. For the solution to the problem with $q \times 1$ objects, we simply put 1 object in each bag.

For $p=43, q=17$, we have $43 = 2 \times 17 + 9$, $17 = 1 \times 9 + 8$, $9 = 1 \times 8 + 1$. Thus our solution is 8 bags with 1 object, 8 bags with 8 objects, 9 bags with 9 objects, 34 bags with 17 objects.