Let us analyze the problem and find the positive integers $n$ such that $\phi(n) \mid n$ and:

$$\sum_{m=1}^{\infty} \left( \left\lfloor \frac{n}{m} \right\rfloor - \left\lfloor \frac{n-1}{m} \right\rfloor \right) = 1992$$

### Step 1: Simplify the Sum

The expression inside the sum, $\left\lfloor \frac{n}{m} \right\rfloor - \left\lfloor \frac{n-1}{m} \right\rfloor$, evaluates to 1 if $m$ divides $n$, and 0 otherwise. Thus, the sum counts the number of divisors of $n$.

This implies:

$$d(n) = 1992$$

where $d(n)$ is the number of divisors of $n$.

### Step 2: Consider $\phi(n) \mid n$

The condition $\phi(n) \mid n$ implies that $n$ must be a power of a prime, say $n = p^k$.

For prime powers, $\phi(n) = p^k - p^{k-1} = p^{k-1}(p-1)$, and divides $n = p^k$.

### Step 3: Prime Power Condition

Using the formula for the number of divisors of a prime power:

$$d(p^k) = k + 1 = 1992$$

Hence, $k = 1991$.

### Step 4: Maximize $n$

To find the largest $n$, assume $p$ is the smallest prime, $p = 3$ (since 2 will not satisfy $\phi(n) \mid n$ for powers greater than 1).

Thus, $n = 3^{1991}$.

$$n = 3^{1991}$$

Hence, the largest $n$ satisfying both $\phi(n) \mid n$ and $d(n) = 1992$ is:

$$\boxed{3^{1991}}$$