$\phi(n)/n$

for a given $k$ , $n=2^{\alpha}3^{\beta}7^{\gamma}$ , for integer $0\leq \alpha,\beta,\gamma \leq k$ .

Then we can consider cases

1- $1\leq \alpha,\beta,\gamma \leq k$

$\frac{\phi(2^{\alpha}3^{\beta}7^{\gamma})}{2^{\alpha}3^{\beta}7^{\gamma}}=\frac{2}{7}$

there is gonna be $k^3$ such numbers $n=2^{\alpha}3^{\beta}7^{\gamma}$

2- exactly one of $\alpha,\beta,\gamma$ is zero. If $\alpha=0$

$\frac{\phi(3^{\beta}7^{\gamma})}{3^{\beta}7^{\gamma}}=\frac{4}{7}$

there is gonna be $k^2$ such numbers $n=2^{\alpha}3^{\beta}7^{\gamma}$

similarly, if $\beta=0$ or If $\gamma=0$ , the fraction would be $\frac{3}{7}$ and $\frac{1}{3}$ , respectively. Each would contain $k^2$ numbers $n$ .

3- exactly two of $\alpha,\beta,\gamma$ are zero. If $\alpha\neq 0$

$\frac{\phi(2^{\alpha})}{2^{\alpha}}=\frac{1}{2}$

there is gonna be $k$ such numbers $n=2^{\alpha}3^{\beta}7^{\gamma}$

similarly, ff $\beta\neq 0$ or If $\gamma\neq 0$ , the fraction would be $\frac{2}{3}$ and $\frac{6}{7}$ , respectively. Again, each would contain $k$ numbers $n$ .

4- all three of $\alpha,\beta,\gamma$ are zero.

$\frac{\phi(1)}{1}=1$

there would be a single number $n=1$ in this case.

So, we form the summation again

$\frac{2}{7}k^3+\frac{28}{21}k^2+\frac{85}{42}k+1=\frac{12k^3+56k^2+85k}{42}+1$

so we need to find the largest $k<1000$ ,for which $\frac{12k^3+56k^2+85k}{42}$ is an integer. From this point, I ran a program to find the largest $k$ , although an analytical approach seems to be easily achievable (rather cumbersome though). the largest would be $k=996$

The function $g(n) = \sum\limits_{d|n} \frac{\varphi(n)}{n}$ is multiplicative , so $g(42^k) = g(2^k)g(3^k)g(7^k).$

For any prime $p$ and nonnegative integer $k,$ direct computation shows that $g(p^k) = 1 + k(1-1/p).$ So $g(42^k) = \left( 1 + \frac{k}2 \right) \left( 1 + \frac{2k}3 \right) \left( 1 + \frac{6k}7 \right) = \frac{(k+2)(2k+3)(6k+7)}{42}.$ This is an integer if and only if the numerator is divisible by $2,3,$ and $7,$ which happens if and only if $\begin{aligned} k &\equiv 0 \pmod 2 \\ k &\equiv 0,1 \pmod 3 \\ k &\equiv 0,2,5 \pmod 7 \end{aligned}$ By the Chinese remainder theorem , this is equivalent to $k \equiv 0, 12, 16, 28, 30, 40 \pmod{42}$ and since $1000 \equiv 34 \pmod{42},$ the largest integer less than $1000$ that fits the requirements is $\fbox{996}.$