Let's determine the smallest positive integer \( a \) such that \( 47^n + a \cdot 15^n \) is divisible by 1984 for all odd \( n \).
If $n$ is odd, then $47^{n}+a \cdot 15^{n}=47\left(47^{n-1}-15^{n-1}\right)+15^{n-1}(47+a \cdot 15)$. Since $n-1$ is even, the first term of the sum is divisible by $\left(47^{2}-15^{2}\right)$, which is exactly 1984. Therefore, $47^{n}+a \cdot 15^{n}$ is divisible by 1984 if and only if $47+a \cdot 15$ is a multiple of 1984. We are looking for the smallest positive integer $a$ such that $47+a \cdot 15$ is divisible by 1984, that is, for which there exists an integer $b$ such that

$$
47+a \cdot 15=b \cdot 1984
$$

If $a$ is positive in (1), then $b$ is also positive, and (1) has minimal positive solutions for $a$ and $b$ simultaneously. Therefore, we need to find the smallest positive integer $b$ for which (1) holds.

Let's examine the numbers involved in terms of divisibility by 15. The left side gives a remainder of 2 when divided by 15, so this must also be true for the right side. Since $b \cdot 1984=b \cdot 1980+b \cdot 4$ and 1980 is divisible by 15, we need to find the smallest positive $b$ for which $4 \cdot b$ gives a remainder of 2 when divided by 15. It is easily verified that this first holds for $b=8$, in which case $b \cdot 4=32=2 \cdot 15+2$.

Substituting $b=8$ into (1) yields $a=1055$, so 1055 is the smallest positive integer $a$ for which $47^{n}+a \cdot 15^{n}$ is divisible by 1984 for all odd $n$.