Any number n can be factorised as 2k *m.
Here m is an odd number because all the 2's in the factors of n is in 2k.
Now,since we are selecting the numbers from {1,2,3,....,200}
So, m can take value in {1,3,5,...,199}
Hence,there are 100 values for m.
Now, we are picking 101 integers from {1,2,...,200}
So, by pigonhole principle we get there exist at least two numbers from the picked numbers such that
m is equal in both the numbers.
Let, the two numbers be k1 and k2
Then k1 = 2p * m
and k2 = 2q * m
Hence k1/k2 = 2p-q
if p<=q then k2 divides k1 otherwise k1 divides k2
So, among the chosen 101 integers there must exist at least two integers such that one is divisible by the other.