Let $ S(n)$ be the sum of decimal digits of a natural number $ n$. Find the least value of $ S(m)$ if $ m$ is an integral multiple of $ 2003$.
1. **Claim 1:**
   - We need to show that \(1001\) is the order of \(10\) modulo \(2003\).
   - The order of an integer \(a\) modulo \(n\) is the smallest positive integer \(k\) such that \(a^k \equiv 1 \pmod{n}\).
   - To prove this, we need to show that \(10^{1001} \equiv 1 \pmod{2003}\) and that no smaller positive integer \(k\) satisfies this condition.

   - **Proof:**
     - Consider the set of all positive divisors of \(1001\). The divisors are \(1, 7, 11, 13, 77, 91, 143, 1001\).
     - We need to check if any of these divisors (other than \(1001\)) is the order of \(10\) modulo \(2003\).
     - After calculations, we find that none of these numbers (except \(1001\)) is an order modulo \(2003\).
     - We verify that \(10^{1001} \equiv 1 \pmod{2003}\) through detailed calculations:
       \[
       10^4 \equiv -15 \pmod{2003}
       \]
       \[
       15^4 \equiv 550 \pmod{2003}
       \]
       \[
       10^{16} \equiv 550 \pmod{2003}
       \]
       \[
       550^2 \equiv 47 \pmod{2003}
       \]
       \[
       10^{32} \equiv 47 \pmod{2003}
       \]
       \[
       10^{64} \equiv 206 \pmod{2003}
       \]
       \[
       10^{96} \equiv 206 \cdot 47 \equiv 1670 \pmod{2003}
       \]
       \[
       10^{100} \equiv 1670 \cdot (-15) \equiv 989 \pmod{2003}
       \]
       \[
       10^{200} \equiv 989^2 \equiv 657 \pmod{2003}
       \]
       \[
       10^{400} \equiv 1004 \pmod{2003}
       \]
       \[
       10^{800} \equiv 507 \pmod{2003}
       \]
       \[
       10^{1000} \equiv 657 \cdot 507 \equiv 601 \pmod{2003}
       \]
       \[
       10^{1001} \equiv 6010 \equiv 1 \pmod{2003}
       \]
     - Therefore, \(1001\) is indeed the order of \(10\) modulo \(2003\).

2. **Claim 2:**
   - There is no number \(n\) with \(S(n) = 2\) that is divisible by \(2003\).

   - **Proof:**
     - Assume the contrary, that there exists \(n\) of the form \(2 \cdot 10^k\) or \(10^k + 1\) that is divisible by \(2003\).
     - The first case \(2 \cdot 10^k\) is clearly impossible because \(2003\) is not divisible by \(2\).
     - For the second case, if \(10^k \equiv -1 \pmod{2003}\), then \(2k\) must be divisible by \(1001\). Thus, \(k\) must be divisible by \(1001\), but then \(10^k \equiv 1 \pmod{2003}\), which is a contradiction.

3. **Claim 3:**
   - There exist natural numbers \(a, b\) such that \(10^a + 10^b + 1\) is divisible by \(2003\).

   - **Proof:**
     - Assume the contrary. Consider the sets of remainders modulo \(2003\):
       \[
       A = \{10^a \mid a = 1, \dots, 1001\}
       \]
       \[
       B = \{-10^b - 1 \mid b = 1, \dots, 1001\}
       \]
     - According to Claim 1, neither of the elements from one set gives the same remainder.
     - Claim 2 implies that neither of these two sets contains \(0\) or \(2002\), so in total, these two sets have at most \(2001\) elements.
     - Therefore, \(A\) and \(B\) must have some common element, meaning:
       \[
       10^a \equiv -10^b - 1 \pmod{2003}
       \]
       \[
       10^a + 10^b + 1 \equiv 0 \pmod{2003}
       \]
     - This number has a sum of digits precisely equal to \(3\).

\(\blacksquare\)

The final answer is \( \boxed{ 3 } \).