The product of two numbers is 4107. If the H.C.F of these two numbers is 37 then the greatest number is

$37a\times 37b=4107$

$ab=\frac{4107}{37\times 37}=3$

Co-primes of with product 3 are $\left(1,3\right)$

$\therefore$ Required numbers are $37\times 1,37\times 3=37,111$