We call a positive integer [i]alternating[/i] if every two consecutive digits in its decimal representation are of different parity.

Find all positive integers $n$ such that $n$ has a multiple which is alternating.
To solve the problem, we need to find all positive integers \( n \) such that \( n \) has a multiple which is alternating. An alternating number is defined as a number where every two consecutive digits have different parity (one is even, the other is odd).

### Case 1: \(\gcd(n, 10) = 1\)
If \( n \) is relatively prime to 10, we can find an alternating multiple of \( n \). Consider the number \( 1010101\ldots101 \), which can be written as:
\[
\frac{10^{2k} - 1}{99}
\]
for some integer \( k \). This number is alternating and we need to show that it is a multiple of \( n \). Since \( \gcd(n, 10) = 1 \), there exists some \( k \) such that \( 99n \mid 10^{2k} - 1 \). Therefore, \( n \) has an alternating multiple.

### Case 2: \( n \) is a power of 5
We will show by induction that there exists an alternating multiple of \( 5^k \) with exactly \( k \) digits.

**Base cases:**
- \( 5 \): The number 5 itself is alternating.
- \( 25 \): The number 25 is alternating.
- \( 125 \): The number 125 is alternating.
- \( 625 \): The number 625 is alternating.
- \( 3125 \): The number 3125 is alternating.

**Inductive step:**
Suppose \( 5^k \) has an alternating multiple with \( k \) digits. We need to show that \( 5^{k+1} \) has an alternating multiple with \( k+1 \) digits. Let \( m \) be the alternating multiple of \( 5^k \). We need to find a digit \( d \) such that:
\[
10^k \cdot d \equiv -m \pmod{5^{k+1}}
\]
Since \( -m \equiv \{0, 5^k, 2 \cdot 5^k, 3 \cdot 5^k, 4 \cdot 5^k\} \pmod{5^{k+1}} \), we can choose \( d \) such that \( 10^k \cdot d \equiv -m \pmod{5^{k+1}} \). This completes the induction.

### Case 3: \( n \) is a power of 5 times something relatively prime to 10
Suppose \( n = 5^k \cdot l \) where \( \gcd(l, 10) = 1 \). If \( k \) is odd, consider an alternating multiple of \( 5^{k+1} \) with \( k+1 \) digits. If \( k \) is even, consider an alternating multiple of \( 5^k \) with \( k \) digits.

We know \( l \) has a multiple of the form:
\[
\frac{10^{kd} - 1}{10^k - 1}
\]
which is alternating. If \( k \) is even, multiply this number by the alternating multiple of \( 5^k \). If \( k \) is odd, replace \( k \) with \( k+1 \) and multiply by \( 5^{k+1} \).

### Case 4: \( n \) is 10 times an odd number
If \( n = 10m \) where \( m \) is odd, take the alternating multiple of \( m \) and place a 0 at the end. This ensures the number is still alternating.

### Case 5: \( n \) is a power of 2
We will show by induction that there exists an alternating multiple of \( 2^k \) with \( k \) digits.

**Base cases:**
- \( 2 \): The number 2 itself is alternating.
- \( 4 \): The number 24 is alternating.
- \( 8 \): The number 248 is alternating.
- \( 16 \): The number 24816 is alternating.

**Inductive step:**
Suppose we have an alternating multiple of \( 2^{k+2} \) with \( k \) digits, where \( k \) is even. We need to show there exists an alternating number \( 2^{k+1} \pmod{2^{k+2}} \) with \( k+1 \) digits. Add an even digit \( d \) such that:
\[
10^k \cdot d \equiv 2^{k+1} \pmod{2^{k+2}}
\]
Clearly, adding 2 or 6 works.

### Case 6: \( n \) is a power of 2 times an odd number
Since the odd number is relatively prime to 10, we can construct a multiple of \( 2^k \) with \( k \) digits. The multiple will still be alternating.

### Conclusion
We have covered all positive integers with \( 20 \nmid n \). Therefore, the answer is:
\[
\boxed{20 \nmid n}
\]