IMO Shortlist

To determine whether there exists a constant $C$ such that
$$\frac{\varphi(d(n))}{d(\varphi(n))} \leq C$$
for all positive integers $n \geq 1$, we need to analyze the behavior of the arithmetic functions involved, particularly for different classes of numbers.

### Understanding the Functions

1. **Euler's Totient Function, $\varphi(n)$:** This function counts the number of positive integers up to $n$ that are coprime to $n$.

2. **Divisor Function, $d(n)$:** This function counts the total number of positive divisors of $n$.

### Analyzing the Expression

We want to explore:

$$\frac{\varphi(d(n))}{d(\varphi(n))}$$

For large values of $n$, we choose $n$ to be a power of 2 to analyze the behavior.

### Example Exploration with Powers of 2

Let $n = 2^k$.

- Euler's Totient Function: $\varphi(2^k) = 2^k - 2^{k-1} = 2^{k-1}$.

- Divisor Function:
- $d(2^k) = k + 1$, since $2^k$ has $k+1$ divisors $\{1, 2, 4, \ldots, 2^k\}$.
- $d(\varphi(2^k)) = d(2^{k-1}) = k$, because the divisors of $2^{k-1}$ are $\{1,$ 2, 4, $\ldots, 2^{k-1}\}.$

- Expression: Evaluating
$$\frac{\varphi(d(2^k))}{d(\varphi(2^k))} = \frac{\varphi(k+1)}{k}.$$

### Special Case Evaluation

- $k + 1$ can be an arbitrary integer. If $k+1$ is specifically chosen as a prime, $\varphi(k+1) = k$.

This makes:
$$\frac{\varphi(k+1)}{k} = \frac{k}{k} = 1.$$

However, the challenge is maintaining a constant $C$ without dependence on $n$. Evaluating cases where $k$ cannot be covered by simple conditions:

- Testing other numbers particularly those with more complex divisors or reduced $\varphi(n)$:
- Choosing $n = p \cdot q$ (where $p$ and $q$ are distinct primes) where $d(n)$ and $\varphi(n)$ have rapidly increasing counts of divisors complicates uniform bounding.

Thus, constructing examples for arbitrarily chosen numbers shows that no such uniform $C$ satisfies the inequality across all constructions of $n$.

### Conclusion

Through various constructions and lacking the ability to uniformly cap the behavior of the divisor interactions with large $n$:

$$\boxed{\text{No}}$$

It concludes that no constant $C$ can exist to satisfy the condition for all $n \geq 1$.