We construct a bijection between possible combinations in Oddland and possible combinations in Squareland. Suppose we have a combination of Squareland stamps that sum to $n$ cent, consisting of $a_1$ stamps of value $1$ cent, $a_2$ stamps of value $4$ cent, ..., $a_M$ stamps of value $M^2$ cent, so that
$$n = \sum_{k=1}^{M} k^2 a_k.$$
Now we express $k^2$ as $\sum_{j=1}^{k}(2j - 1)$ and interchange the order of summation, which yields
$$n = \sum_{k=1}^{M} \sum_{j=1}^{k} (2j - 1) a_k = \sum_{j=1}^{M} (2j - 1) \sum_{k=j}^{M} a_k.$$
This gives us a possible combination of Oddland stamps: By setting $b_j = \sum_{k=j}^{M} a_k$, we have
$$n = \sum_{j=1}^{M} (2j - 1)b_j.$$
This can be interpreted as a collection of $b_1$ stamps of value $1$ cent, $b_2$ stamps of value $3$ cent, ..., $b_M$ stamps of value $(2M - 1)$ cent. We have $b_1 \ge b_2 \ge \dots \ge b_M$ by definition, so this is a legal combination in Oddland.

Conversely, if a combination in Oddland is given by the values $b_1, b_2, \dots, b_M$, we can use the identities $a_1 = b_1 - b_2, a_2 = b_2 - b_3, \dots, a_{M-1} = b_{M-1} - b_M, a_M = b_M$ to recover the corresponding combination in Squareland. (Note that these values are nonnegative whenever $b_1 \ge b_2 \ge \dots \ge b_M$.)

Since these two operations obviously are inverse to one another, we have found a bijection, which proves the statement.