To find the smallest possible value of $n$ that has exactly 16 different positive integer divisors, we need to consider the number of divisors function. For a number $n$ with prime factorization $n = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k}$, the number of divisors $d(n)$ is given by:

$$d(n) = (e_1 + 1)(e_2 + 1) \cdots (e_k + 1)$$

We need $d(n) = 16$. The possible factorizations of 16 are:

1. $16 = 16 \times 1$
2. $16 = 8 \times 2$
3. $16 = 4 \times 4$
4. $16 = 4 \times 2 \times 2$
5. $16 = 2 \times 2 \times 2 \times 2$

We will consider each case to find the smallest $n$.

Case 1: $n = p_1^{15}$
- The smallest prime is 2.
- $n = 2^{15} = 32768$

Case 2: $n = p_1^7 \cdot p_2$
- The smallest primes are 2 and 3.
- $n = 2^7 \cdot 3 = 128 \cdot 3 = 384$

Case 3: $n = p_1^3 \cdot p_2^3$
- The smallest primes are 2 and 3.
- $n = 2^3 \cdot 3^3 = 8 \cdot 27 = 216$

Case 4: $n = p_1^3 \cdot p_2 \cdot p_3$
- The smallest primes are 2, 3, and 5.
- $n = 2^3 \cdot 3 \cdot 5 = 8 \cdot 3 \cdot 5 = 120$

Case 5: $n = p_1 \cdot p_2 \cdot p_3 \cdot p_4$
- The smallest primes are 2, 3, 5, and 7.
- $n = 2 \cdot 3 \cdot 5 \cdot 7 = 210$

Among these cases, the smallest value of $n$ is found in Case 4:

$$n = 2^3 \cdot 3 \cdot 5 = 120$$

Thus, the smallest possible value of $n$ that has exactly 16 different positive integer divisors is $\boxed{120}$.