This splits into two subcases: n is a perfect square or not.

If n is a perfect square then let m=sqrt(n).  Then m is an integer from 1 to 44.

Then one root is 45-m.  I will choose the other root to be m+2.  Now the quadratic equation becomes x^2-47x+(m+2)*(45-m).

m+2 and 45-m are both less that 47, but since 47 is prime then both are coprime to 47, which makes the coefficients of the quadratic coprime, satisfying that requirement.

44 values on n so far.

Now if n is not a perfect square then we must have a conjugate pair of roots: 45-sqrt(n) and 45+sqrt(n).  Then the quadratic must be x^2-90x+2025-n.

Now for the coprime condition, we must have 90 be coprime to 2025-n.  If n is odd then there is a common factor of 2.  If n has 3 or 5 as a factor then that value is a common factor.  So that leaves us looking for even numbers coprime to 15 on the interval [1,2025].

In any block of 30 values there are 8 such values, the numbers congruent to 2,4,8,14,16,22,26,28 mod 30.  So then over [1,2025] there are 67 blocks plus a half block, which do add up to 67*8+4=540 values of n.

But we have double counted perfect squares.  Those squares would be any even square that is coprime to 15, and there are 12 of those: the squares of 2,4,8,14,16,22,26,28,32,34,38, and 44.

Then for a final answer we have 44+540-12 = 572 values of n that satisfy the problem.