Positive integers

If 2003 is divided by n, the result would be a whole number 'N' plus a remainder.

If the remainder is 23, then,

$\frac{2003}{n}$ = N + $\frac{23}{n}$

or, N = $\frac{1980}{n}$

or, n = $\frac{1980}{N}$

The factors of 1980,

1, 2, 3, 4, 5, 6, 9, 10, 11, 12, 15, 18, 20, 22, 30, 33, 36, 44, 45, 55, 60, 66, 90, 99, 110, 132, 165, 180, 198, 220, 330, 396, 495, 660, 990 and 1980

n must be greater than 23 [ otherwise the remainder cannot be 23 ]

Therefore, there are 22 possible values of n = 30, 33, 36, 44, 45, 55, 60, 66, 90, 99, 110, 132, 165, 180, 198, 220, 330, 396, 495, 660, 990 and 1980

Answer : 22

$2003 \equiv 23 \pmod{n} \\ \implies 2003 - 23 \equiv 0 \pmod{n} \\ \implies 1980 \equiv 0 \pmod{n}$

Thus $n$ should be a divisor of $1980$ . Also note that $n > 23$ to give a remainder of $23$ when it divides $2003$ .

Prime factorization of $1980$ gives $1980 = 2^2 \times 3^2 \times 5 \times 11$ . Thus $1980$ has $36$ divisors. Listing out the divisors, we find that there are $14$ divisors which are less than $23$ , so our required answer is $36 - 14 = \boxed{22}$ .