, 2024

Evaluating, we get $f(132)=132+1+3+2=138$.
Suppose that $n$ is equal to
the 3-digit positive integer $abc$.

Then $f(n)=f(abc)=100a+10b+c+a+b+c=101a+11b+2c$.

Since $f(n)=175$, then $101a+11b+2c=175$.

It cannot be the case that $a $\geq$
2$, since if we had$a $\geq$
2$, then$101a $\geq$ 202$
which is too large, noting that $11b+2c$ is always at least $0$.

Therefore, $a < 2$ which means
that $a = 1$.

When $a = 1$, we get $101+11b+2c=175$ or $11b+2c=74$.

It cannot be the case that $b $\geq$
7$, since if we had$b $\geq$
7$, then$11b $\geq$ 77$ which
is too large, noting that $2c$ is
always at least $0$.

Therefore, $b < 7$. If $b=6$, then $66+2c=74$ or $2c=8$, and so $c=4$.

If $b\leq5$, then $11b\leq55$, and so $2c\geq74-55=19$, which is not possible
since $c\leq9$.

We can confirm that $f(164)=164+1+6+4=175$, and so $n=164$.
Suppose that $n$ is equal to
the 3-digit positive integer $pqr$.

Then $f(pqr)=100p+10q+r+p+q+r$, and
so $101p+11q+2r=204$.

If $p\geq3$, then $101p\geq303$, and so $p=1$ or $p=2$.

If $p=1$, then $101+11q+2r=204$ or $11q+2r=103$.

Since $r \leq 9$, then $2r \leq 18$ and so $11q \geq 103 - 18 = 85$.

Therefore, $q=8$ or $q=9$.

If $q=8$, then $88+2r=103$ or $2r=15$, which is not possible since $r$ is an integer.

If $q=9$, then $99+2r=103$ or $2r=4$, and so $r=2$.

In this case, $n=192$ and we can
confirm that $f(192)=192+1+9+2=204$.

If $p=2$, then $202+11q+2r=204$ or $11q+2r=2$.

The only possible solution to $11q+2r=2$ is $q=0$ and $r=1$.

In this case, $n=201$ and we can
confirm that $f(201)=201+2+0+1=204$.

Therefore, if $f(n)=204$, then the
possible values of $n$ are $192$ and $201$.