Let's find numbers that give a remainder of 15 when their product with 37 is divided by 31.
If there is such a number $x$, then according to the problem,

$$
37 x=31 y+15
$$

from which

$$
y=x+\frac{6 x-15}{31}=x+u
$$

so

$$
31 u=6 x-15
$$

from which

$$
x=5 u+2+\frac{u+3}{6}=5 u+2+v
$$

thus

$$
6 v=u+3
$$

and

$$
u=6 v-3
$$

Therefore,

$$
x=31 v-13
$$

By substituting any positive integer for $v$, we get numbers that satisfy the conditions. For example, 18, 49, 80, etc.

(Jenő Engler, Pécs.)

Number of solutions: 33.