Problem 1437

1. Define Partial Sums:
Consider the partial sums $s_i = a_1 + a_2 + \ldots + a_i$ for $i = 1, 2, \ldots, 51$. Clearly, $s_1, s_2, \ldots, s_{51}$ are natural numbers ranging from 1 to 100, inclusive, since the sum of all 51 numbers is 100.

2. **Case 1: $k \geq 50$:**
- Create pairs $(1, k+1), (2, k+2), \ldots, (100 - k, 100)$. There are $100 - k \leq 50$ such pairs.
- By the Pigeonhole Principle, since we have 51 partial sums and only $100 - k$ pairs, at least one pair must receive two partial sums. Suppose pair $(j, j+k)$ receives partial sums $s_{\ell}$ and $s_m$ with $\ell < m$.
- Then, $s_m - s_{\ell} = a_{\ell + 1} + a_{\ell + 2} + \ldots + a_m = (j+k) - j = k$.
- Thus, there exist consecutive terms whose sum is $k$.

3. **Case 2: $k < 50$:**
- Create pairs $(1, 101 - k), (2, 102 - k), \ldots, (k, 100)$. There are $k < 50$ such pairs.
- By the Pigeonhole Principle, since we have 51 partial sums and only $k$ pairs, at least one pair must receive two partial sums. Suppose pair $(j, j + (100 - k))$ receives partial sums $s_{\ell}$ and $s_m$ with $\ell < m$.
- Then, $s_m - s_{\ell} = a_{\ell + 1} + a_{\ell + 2} + \ldots + a_m = (j + (100 - k)) - j = 100 - k$.
- Thus, there exist consecutive terms whose sum is $100 - k$.

4. Conclusion:
In either case, for any natural number $1 \leq k < 100$, there are some consecutive numbers from the sequence whose sum is either $k$ or $100 - k$.

$\blacksquare$