Combinatorics problem (Pigeonhole principle).
Solution 1:
Consider the sequence of initial sums $S(0),S(1),\ldots,S(55)$ $$ S(n)=\sum_{i=1}^na_i. $$ We have $$S(0)=0<S(1)<S(2)<\cdots<S(55)<95,$$ a sequence of 56 distinct integers in the interval $[0,94]$ of $95$ possibilities.
Consider also the sequence of numbers $T(i)=S(i)-15$. There are 56 of those as well, all integers in the range $[-15,79]$. Because they are all distinct, at least $56-15=41$ of them are non-negative.
Hints:
- Can you show that the sets $\{S(i)\mid 0\le i<56\}$ and $\{T(i)\mid 0\le i<56\}$ must have a non-empty intersection?
- Do you see how the claim follows from this?
Spoiler #1
Between them the two sets have 97 positive integers in the range $[0,94]$ so the pigeonhole principle tells us that there is some overlap.
Spoiler #2
If $T(\ell)=S(\ell)-15=S(k)$ then $$ \sum_{i=k+1}^\ell a_i=S(\ell)-S(k)=15.$$
Solution 2:
This is essentially the same answer as already given, but from a slightly different point of view. Keep this trick, because it pops up all over the place, for instance in the proof of Chebyshev's Inequality.
Instead of writing $$ S = \sum_{i=1}^{55} a_i $$ where $a_i \in \{1, 2, 3, \cdots\}$, let $N_k$ be the number of $i \in \{1, \cdots, 55\}$ such that $a_i = k$ and write $$ S = \sum_{k=1}^\infty k \cdot N_k. $$ That is, group the $a_i$ by value and add up the totals for each group. Within a group, each element has the same value so you need only count the elements. If you're familiar with measure theory, this is essentially converting a Riemann integral into a Lebesgue integral. (Conversely, if measure theory is confusing, thinking of it like this might help.)
Now observe that for any $a_i > 1$, we are adding at least $2$ to the sum, so we can approximate like this: \begin{align} S &= N_1 + \sum_{k=2}^\infty k \cdot N_k \\ &\ge N_1 + 2 \cdot \sum_{k=2}^\infty N_k \\ &= N_1 + 2 \cdot (55 - N_1) \\ &= 110 - N_1. \end{align}
From here, use the fact that $S < 95$ to show that there are a large number of $a_i$ with $a_i = 1$.