Problem 810

Answer: 63, 106, 228, 291, or 673.

Solution. Suppose that for some $n$ there exist $n$ consecutive natural numbers $a$, $a+1, \ldots, a+(n-1)$, the sum of which is 2016. Then $n a + (n-1) n / 2 = 2016$, or, after algebraic transformations, $n(2 a + n - 1) = 4032 = 2^{6} \cdot 3^{2} \cdot 7$.

Note that $n$ and $2 a + n - 1$ have different parities. Therefore, if $n$ is even, then $n$ must be divisible by $2^{6} = 64$, hence $n \geq 64, 2 a + n - 1 \leq 4032 / 64 = 63$, which contradicts $2 a + n - 1 > n$.

Thus, $n$ is an odd divisor of the number 4032, i.e., a divisor of the number 63. Let's check that each of them works and find the corresponding value of $a + (n-1)$. If $n=3: 2 a + 3 - 1 = 1344, a = 671$, $a + (n-1) = 673$; if $n=7: 2 a + 7 - 1 = 576, a = 285, a + (n-1) = 291$; if $n=9: 2 a + 9 - 1 = 448$, $a = 220, a + (n-1) = 228$; if $n=21: 2 a + 21 - 1 = 192, a = 86, a + (n-1) = 106$; if $n=63$: $2 a + 63 - 1 = 64, a = 1, a + (n-1) = 63$.