IMO2024 Shortlisted Problems

Answer: The minimum value is $\frac{n(n+1)}{2}$.

For a fixed $n$, let $f(n)$ denote the minimum possible value of $S$. Consider the following variant: among all infinite sequences of nonnegative integers $x_0,x_1,\dots$, only finitely many of which are nonzero, satisfying $x_0+x_1+\cdots =n$, let $g(n)$ denote the minimum possible value of $$T=2^0{x}_0^2+2^1{x}_1^2+2^2{x}_2^2+\cdots$$ It is clear that $g(n)⩽f(n)$. Conversely, it is easy to see that if a sequence $x_0,x_1,\dots$ achieves the minimum of $g(n)$, then $x_0⩾x_1⩾\cdots$ and thus $x_{n+1}=x_{n+2}=\cdots =0$. In particular, $f(n)=g(n)$.

Now, we hope to get an inductive formula for $g(n)$. Note that, in order to minimise $T$ for $n⩾1$, we must have $x_0⩾1$ since the sequence ($x_i$) is nonincreasing. Note that the minimal value of $$2^1{x}_1^2+2^2{x}_2^2+\cdots =2\left(2^0{x}_1^2+2^1{x}_2^2+\cdots \right)$$ over all infinite sequences of nonnegative integers with $x_1+x_2+\cdots =m$ is exactly $2g(m)$. As a result, for $n⩾1$ we have $$g(n)=\underset{x_0\in \{1,2,\dots ,n\}}{\min }\left(x_0^2+2g\left(n-x_0\right)\right).$$ We now prove $g(n)=\frac{n(n+1)}{2}$ by induction. It is clear that $g(0)=0$. Assume that this has been proved for $n=0,1,\dots ,N-1$. Then, $$\begin{array}{rlr}{x}_0^2+2g\left(N-x_0\right)& =x_0^2+\left(N-x_0\right)\left(N-x_0+1\right)& =2x_0^2-(2N+1)x_0+N(N+1)\\ & =\frac{1}{2}\left[\left(2x_0-N\right)\left(2x_0-N-1\right)+N^2+N\right]& \end{array}$$ The product of two consecutive integers $\left(2x_0-N\right)\left(2x_0-N-1\right)$ is always nonnegative, and it is zero precisely when $2x_0$ is the even number in $\{N,N+1\}$. Thus the minimum of the final expression in equation (1) is $\frac{1}{2}\left(N^2+N\right)$, so $g(N)=\frac{N(N+1)}{2}$, completing the inductive proof.

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Alternative solution.

Consider the following table of numbers, where the row and column indices start from $0$, and $a_{i,j}=2^i(2j+1)$ for $i,j⩾0$.

	$j=0$	1	2	3	4	5	$\cdots$
$i=0$	1	3	5	7	9	11
1	2	6	10	14	18	22
2	4	12	20	28	36	44
3	8	24	40	56	72	88
4	16	48	80	112	144	176
$⋮$

Every number can be written uniquely as a product of a power of $2$ and an odd number so every positive integer appears exactly once in the table above. It is easy to see that numbers in each row and each column are strictly increasing. Since the sum of the first $x$ odd positive integers is $x^2$, the sum of the first $x_k$ numbers in the $k^{\text{th}}$ row is $2^k{x}_k^2$, the $k^{\text{th}}$ term appearing in $S$.

Thus, the sum $S$ can be interpreted as the result of taking a total of $n$ numbers from the first $n$ rows of the table such that we take the leftmost $x_k$ numbers from row $k$ (where ${\sum }_{k=1}^n{x}_k=n$), and then computing the sum of these $n$ numbers. In particular, the minimum possible value of $S$ is the same as the sum of the smallest $n$ numbers in this table, since every row and every column of the table is strictly increasing.

Moreover, the smallest $n$ numbers, namely $1,2,\dots ,n$, appear in the first $n$ rows, so the minimum of $S$ is $$1+2+\cdots +n=\frac{n(n+1)}{2}.$$