USAMO

Solution 1
Let one optimal set of integers be $\{a_1,\dots,a_{2k+1}\}$ with $a_1 > a_2 > \cdots > a_{2k+1} > 0$ .
The two conditions can now be rewritten as $a_1+\cdots + a_k \leq N/2$ and $a_1+\cdots +a_{2k+1} > N$ .
Subtracting, we get that $a_{k+1}+\cdots + a_{2k+1} > N/2$ , and hence $a_{k+1}+\cdots + a_{2k+1} > a_1+\cdots + a_k$ .
In words, the sum of the $k+1$ smallest numbers must exceed the sum of the $k$ largest ones.
Let $a_{k+1}=C$ . As all the numbers are distinct integers, we must have $\forall i \in\{1,\dots,k\}:~ a_{k+1-i} \geq C+i$ , and also $\forall i \in\{1,\dots,k\}:~ a_{k+1+i} \leq C-i$ .
Thus we get that $a_1+\cdots + a_k \geq kC + \dfrac{k(k+1)}2$ , and $a_{k+1}+\cdots + a_{2k+1} \leq (k+1)C - \dfrac{k(k+1)}2$ .
As we want the second sum to be larger, clearly we must have $(k+1)C - \dfrac{k(k+1)}2 > kC + \dfrac{k(k+1)}2$ .
This simplifies to $C > k(k+1)$ .
Hence we get that:
$$\begin{align*} N & \geq 2(a_1+\cdots + a_k) \\ & \geq 2\left( kC + \dfrac{k(k+1)}2 \right) \\ & = 2kC + k(k+1) \\ & \geq 2k(k^2+k+1) + k(k+1) \\ & = 2k^3 + 3k^2 + 3k \end{align*}$$
On the other hand, for the set $\{ k^2+k+1+i ~|~ i\in\{-k,\dots,k\} \, \}$ the sum of the largest $k$ elements is exactly $k^3 + k^2 + k + \dfrac{k(k+1)}2$ , and the sum of the entire set is $(k^2+k+1)(2k+1) = 2k^3 + 3k^2 + 3k + 1$ , which is more than twice the sum of the largest set.
Hence the smallest possible $N$ is $\boxed{ N = 2k^3 + 3k^2 + 3k }$ .
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.