An engineer designs at least one robot a day for 30 days. If a total of 45 robots have been designed, then show that there must have been a series of consecutive days when exactly 14 robots were designed.

Let $a_i$ be the number of robots designed on the $i$th day. Then we have 30 numbers $a_1,a_2,…,a_{30}$ and $a_i\geq 1$ for all $1\leq i\leq 30$ .

Let $s_k=\sum_{i=1}^ka_i$ , it is the number of robots designed on or before the $k$th day.

We have a sequence of 30 numbers $s_1,s_2,…,s_{30},\ \text{where }$ $1\leq s_1<s_2<…<s_{30}=45$ .

Let us consider new sequence: $s_1+14, \ s_2+14,\ …,\ s_{30}+14$ . There are 30 numbers and $s_1+14<s_2+14<…<s_{30}+14=45+14=59$ .

There are totally 60 numbers: $s_1,s_2,…,s_{30},s_1+14,s_2+14,…,s_{30}+14$ and all of them are less or equal than $59$ ($s_i\leq 45$ and $s_i+14\leq 59$ ).

By the Pigeonhole Principle, at least two of these numbers are equal.

Since $s_i\neq s_j$ and $s_i+14\neq s_j+14$ for all $i\neq j$, it follows that $s_i=s_j+14$ for some $i$ and $j$ .

Now we can conclude that exactly 14 robots were designed from day $j+1$ to day $i$.