Problem 3. For which natural numbers $n>1$ will there be $n$ consecutive natural numbers whose sum is equal to $2016?$
Answer: For $n$ equal to $3, 7, 9, 21$ or 63.

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, $n(2a + n - 1) = 4032 = 2^6 \cdot 3^2 \cdot 7$.

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

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. If $n=3: 2a + 3 - 1 = 1344, a = 671$; if $n=7: 2a + 7 - 1 = 576, a = 285$; if $n=9$: $2a + 9 - 1 = 448, a = 220$; if $n=21: 2a + 21 - 1 = 192, a = 86$; if $n=63: 2a + 63 - 1 = 64, a = 1$.