USA TST

Replace $300 = \frac{24 \cdot 25}{2}$ with $\frac{s(s+1)}{2}$ where $s = 24$, and 20 with $k$. The answer is $115440 = \frac{ks(ks+1)}{2}$. Equality is achieved at $f(n) = \max(s-n, 0)$ and $g(n) = \max(ks-n, 0)$. To prove
$$g(n_1 + \dots + n_k) \le f(n_1) + \dots + f(n_k),$$
write it as
$$\max(x_1 + \dots + x_k, 0) \le \max(x_1, 0) + \dots + \max(x_k, 0)$$
with $x_i = s - n_i$. This can be proven from the $k=2$ case and induction.

It remains to show the upper bound. For this problem, define a partition to be a nonincreasing function $p: \mathbb{Z}_{\ge 0} \to \mathbb{Z}_{\ge 0}$ such that $p(n) = 0$ for some $n$. The sum of $p$ is defined to be $\sum_{n=0}^{\infty} p(n)$, which is finite under the previous assumption. Let $L = \mathbb{Z}_{\ge 0}^2$. The Young diagram of the partition is the set of points
$$\mathcal{P} := \{(x, y) \in L : y < p(x)\}.$$
The number of points in $\mathcal{P}$ is equal to the sum of $p$. The conjugate of a partition defined as
$$p_*(n) = \text{the number of } i \text{ for which } p(i) > n.$$
This is a partition with the same sum as $p$. Geometrically, the Young diagrams of $p$ and $p_*$ are reflections about $x = y$.
Since each $g(n)$ is independent, we may maximize each one separately for all $n$ and assume that
$$g(n) = \min_{n_1+\cdots+n_k=n} (f(n_1) + \cdots + f(n_k)). \quad (*)$$
The conditions of the problem statement imply that $f(\frac{s(s+1)}{2}) = 0$. Then, for any $n \le k \frac{s(s+1)}{2}$, there exists an optimal combination $(n_1, \dots, n_k)$ in (*) where all $n_i$ are at most $\frac{s(s+1)}{2}$, by replacing any term in an optimum greater than $\frac{s(s+1)}{2}$ by $\frac{s(s+1)}{2}$ and shifting the excess to smaller terms (because $f$ is nonincreasing). Therefore we may extend $f$ to a partition by letting $f(n) = 0$ for $n > \frac{s(s+1)}{2}$ without affecting the relevant values of $g$. Then (*) implies that $g$ is a partition as well.

The problem can be restated as follows: $f$ is a partition with sum $\frac{s(s+1)}{2}$, and $g$ is a partition defined by (*). Find the maximum possible sum of $g$. The key claim is that the problem is the same under conjugation.
Claim — Under these conditions, we have
$$g_*(n) = \min_{n_1+\cdots+n_k=n} (f_*(n_1) + \cdots + f_*(n_k)).$$
Proof. Let $\mathcal{F}$ and $\mathcal{G}$ be the Young diagrams of $f$ and $g$ respectively, and $\overline{\mathcal{F}} = L \setminus \mathcal{F}$ and $\overline{\mathcal{G}} = L \setminus \mathcal{G}$ be their complements. The lower boundary of $\overline{\mathcal{F}}$ is formed by the points $(n, f(n))$ for $i \in \mathbb{Z}_{\ge 0}$. By the definition of $g$, the lower boundary of $\overline{\mathcal{G}}$ consists of points $(n, g(n))$ which are formed by adding $k$ points of $\overline{\mathcal{F}}$. This means
$$\overline{\mathcal{G}} = \underbrace{\overline{\mathcal{F}} + \cdots + \overline{\mathcal{F}}}_{k \text{ times}}$$
where $+$ denotes set addition. This definition remains invariant under reflection about $x = y$, which swaps $f$ and $g$ with their conjugates. $\square$

Let $A$ be the sum of $g$. We now derive different bounds on $A$. First, by Hermite's identity
$$n = \sum_{i=0}^{k-1} \left\lfloor \frac{n+i}{k} \right\rfloor$$
we have
$$\begin{align*} A &= \sum_{n=0}^{\infty} g(n) \\ &\le \sum_{n=0}^{\infty} \sum_{i=0}^{k-1} f\left(\left\lfloor \frac{n+i}{k} \right\rfloor\right) \\ &= k^2 \sum_{n=0}^{\infty} f(n) - \frac{k(k-1)}{2} f(0) \\ &= k^2 \frac{s(s+1)}{2} - \frac{k(k-1)}{2} f(0). \end{align*}$$

By the claim, we also get the second bound $A \le k^2 \frac{s(s+1)}{2} - \frac{k(k-1)}{2} f_*(0)$.
For the third bound, note that $f(f_*(0)) = 0$ and thus $g(kf_*(0)) = 0$. Moreover,
$$g(qf_*(0) + r) \le q \cdot f(f_*(0)) + (k - q - 1)f(0) + f(r) = (k - q - 1)f(0) + f(r),$$
so we have
$$\begin{align*} A &= \sum_{\substack{0 \le q < k \\ 0 \le r < f_*(0)}} g(qf_*(0) + r) \\ &\le \frac{k(k-1)}{2} f_*(0) f(0) + k \sum_{0 \le r < f_*(0)} f(r) \\ &= \frac{k(k-1)}{2} f_*(0) f(0) + k \frac{s(s+1)}{2}. \end{align*}$$

* If $f(0) \ge s$ then
$$A \le k^2 \frac{s(s+1)}{2} - \frac{k(k-1)}{2} f(0) \le \frac{ks(ks+1)}{2}.$$
* If $f_*(0) \ge s$ then
$$A \le k^2 \frac{s(s+1)}{2} - \frac{k(k-1)}{2} f_*(0) \le \frac{ks(ks+1)}{2}.$$
* Otherwise, $f(0)f_*(0) \le s^2$ and
$$A \le \frac{k(k-1)}{2} f_*(0) f(0) + k \frac{s(s+1)}{2} \le \frac{ks(ks+1)}{2}.$$

In all cases, $A \le \frac{ks(ks+1)}{2}$, as desired.

Remark. One can estimate the answer to be around $k^2 \frac{s(s+1)}{2}$ by observing the set addition operation “dilates” $\mathcal{F}$ by a factor of $k$, but significant care is needed to sharpen the bound.