$n$ cannot be even, since the product of an even number of odd numbers is odd, while their sum is even.

Let $n$ be odd. Suppose there exist odd numbers that satisfy the conditions, denoted by $2a_{1}+1$, $2a_{2}+1, \ldots, 2a_{n}+1$. Then

$$
\left(2a_{1}+1\right)\left(2a_{2}+1\right) \cdot \ldots \cdot\left(2a_{n}+1\right)=2a_{1}+1+\ldots+2a_{n}+1=2a_{1}+2a_{2}+\ldots+2a_{n}+n
$$

Perform the multiplication, then the left side becomes

$$
\left(1+2a_{1}+2a_{2}+\ldots+2a_{n}\right)+\left(2a_{1}\right)\left(2a_{2}\right)+\left(2a_{1}\right)\left(2a_{3}\right)+\ldots+\left(2a_{1}\right)\left(2a_{2}\right) \cdot \ldots \cdot\left(2a_{n}\right)
$$

In this sum, apart from the $n+1$ terms in the first parentheses, we can be sure that the rest are divisible by 4, so their sum can be written in the form $4k$. Using this, if we subtract $\left(2a_{1}+2a_{2}+\ldots+2a_{n}\right)$ from both sides of (1),

$$
4k+1=n
$$

results. Thus, $n$ can only be of the form $4k+1$.

For any number $n=4k+1$ (where $k \geq 1$), we show a suitable number $n$. Take the numbers

$$
2k+1,3, \underbrace{1,1 \ldots, 1}_{4k-1 \text{ times}}
$$

Their product and sum are both $6k+3$.

Therefore, the answer to the problem is: for numbers $n=4k+1$.

Based on the work of László Székelyhidi (Debrecen, Tóth Árpád Gymnasium, 1st year).
For which $n$ can we find $n$ positive odd numbers whose sum and product are equal?