, 2026

The sum of the first $2k$
positive integers is $$1+2+3+\cdots+(2k-1)+2k = \dfrac{2k(2k+1)}{2}=\dfrac{4k^2+2k}{2}$$ The sum of the $k$ consecutive integers starting at $m\geq 1$ is $$\begin{align*} m + (m+1) + (m+2) + \cdots + (m+k-1) &= km + (1+2+3+\cdots+(k-1)) \\ &= km + \dfrac{(k-1)k}{2} \\ &= \dfrac{k^2+2km-k}{2}\end{align*}$$ Note that we must have
$m\leq k+1$, otherwise $m+k-1>2k$. Thus, the sum of the
remaining integers is $$\dfrac{4k^2+2k}{2} - \left(\dfrac{k^2+2km-k}{2}\right) = \dfrac{3k^2-2km+3k}{2}$$ So,
we are looking for positive integer pairs $(k,m)$ for which $1\leq m \leq k+1$ and $$\dfrac{3k^2-2km+3k}{2} = 819$$ Rearranging
and factoring, this equation is $$k(3k-2m+3) = 1638$$ Thus, we have that $(k,3k-2m+3)$ is a divisor pair of $1638$. We are assuming that $m\leq k+1$, and so $m < k+2$, from which it follows that
$-2m > -2k-4$. Therefore, we have
$$\begin{align*} (3k-2m+3) - k &= 2k-2m+3 \\ &> 2k+(-2k-4) + 3 \\ &= -1 \\\end{align*}$$ Note that for an integer to be
greater than $-1$, it must be at
least $0$. Therefore, $(3k-2m+3) \geq k$, and so in each divisor
pair, $(k,3k-m+3)$, $3k-m+3$ is the greater of the two. ($1638$ is not a perfect square, so the
divisors will never be equal.) The ordered divisor pairs of
$1638$ are

$(1,1638)$, $(7,234)$, $(18,91)$, $(2,819)$, $(9,182)$, $(21,78)$, $(3,546)$, $(13,126)$, $(26,63)$, $(6,273)$, $(14,117)$, $(39,42)$

Now observe that $3(k+1) - (3k-2m+3) = 2m
> 0$. Therefore, if we triple one more than the smaller
divisor, the result must exceed the larger divisor. Of the divisor pairs
above, only $(26,63)$ and $(39,42)$ have this property.

If $k=26$ and $3k-2m+3=63$, then $m=9$, and if $k=39$ and $3k-2m+3=42$, then $m=39$.

The only two possible values of $k$ are $k=26$ and $k=39$, and their sum is $26+39=65$.