For each positive integers $u$ and $n$, say that $u$ is a [i]friend[/i] of $n$ if and only if there exists a positive integer $N$ that is a multiple of $n$ and the sum of digits of $N$ (in base 10) is equal to $u$. Determine all positive integers $n$ that only finitely many positive integers are not a friend of $n$.
1. **Claim**: \( n \) works if and only if \( 3 \nmid n \).

2. **Proof for \( 3 \mid n \)**:
   - If \( 3 \mid n \), then all friends of \( n \) must be multiples of 3. This is because if \( 3 \mid n \) and \( N \) is a multiple of \( n \), then \( N = \overline{d_k \cdots d_0}_{10} \) implies \( \sum d_j \equiv \sum d_j 10^j = N \equiv 0 \pmod{3} \). Therefore, the sum of the digits of \( N \) must also be a multiple of 3.

3. **Proof for \( \gcd(3, n) = 1 \)**:
   - Let \( s(a) \) denote the sum of digits of \( a \) in base 10.
   - If \( u \) and \( v \) are friends of \( n \), then \( u + v \) is also a friend. This is because if \( n \mid a \) and \( s(a) = u \), and \( n \mid b \) and \( s(b) = v \), then \( n \mid (a \cdot 10^N + b) \) for a sufficiently large integer \( N \). Moreover, \( s(a \cdot 10^N + b) = s(a) + s(b) = u + v \).

4. **Application of Chicken McNugget Theorem**:
   - By the Chicken McNugget theorem, if we can find three friends of \( n \) that do not share a prime factor, then we are done. It suffices to find two friends of \( n \) that differ by 9.

5. **Finding Friends Differing by 9**:
   - Let \( n = 10^t n' \) where \( 10 \nmid n' \). Let \( N \) be a sufficiently large number, and \( 9 \cdot 10^N < b < 10^{N+1} \) for a sufficiently large \( b \) where \( n \mid b \). Such \( b \) must exist if \( 10^N > b \).

6. **Sum of Digits Analysis**:
   - Consider \( s(c \cdot 10^N + b) \) where the units digit of \( c \) is not 0. The \( j \)-th digit from the right (where the 0th digit represents the unit digit) of \( c \cdot 10^N + b \) is the same as the \( j \)-th digit of \( b \) if \( j < N \). The \( N \)-th digit of \( c \) has decreased by 1. If the 1st digit of \( c \) is not equal to 9, then the \( (N+1) \)-th digit of \( c \cdot 10^N + b \) is one plus the first digit of \( c \). The \( j \)-th digit of \( c \) is equal to the \( (N+j) \)-th digit of \( c \cdot 10^N + b \).

7. **Conclusion**:
   - It suffices to find a multiple of \( n' \) (call it \( M \)) such that the tens digit of \( M \) is not equal to 9, but the units digit of \( M \) is nonzero. If \( n' \)'s tens digit is not equal to 9, we are done. Otherwise, the tens digit of \( 11n' \) is congruent to the tens digit of \( n' \) plus the units digit of \( n' \) mod 10, which is not 9.

\(\blacksquare\)

The final answer is \( \boxed{ n } \) works if and only if \( 3 \nmid n \).