Choose any different $38$ natural numbers less than $1000$.Prove among the selected numbers there exists at least two whose difference is at most $26$

Question

Choose any different $38$ natural numbers less than $1000$. Prove by using the Pigeonhole Principle that among the selected numbers there exists at least two whose difference is at most $26$.

I proved an answer by increasing $26$ to $27$ and multiplying it by $38$. This gives an answer of $1026$, but I can't use the pigeon principle to show it.

Answer 1

Create $37$ pigeonholes of size $27$ - and note that $27 \times 37 =999$. Then there must be two of the $38$ numbers in the same pigeonhole. Fill in the details for a proof.

Answer 2

RTP: there are at least two pairs whose difference is a maximum of $26$

The difference is maximized when $\frac{1000}{38} \approx 26.315$ so let the set of possible differences be $S_{\text{differences}} = {\{1,2,3,...,26\}}$ .

There are $38$ numbers and only $26$ differences when maximised, so the ceiling function of $\frac{38}{26} =2$

Therefore by PHP, there exists at least two numbers whose difference is at most $26$