Where can we start bounding?

The only such numbers are $1$ , $22$ , $23$ , and $24$ .

$1+22+23+24=\boxed{70}$

This is because you can enumerate all the ones from 1 to 25 and see that 1, 22, 23 and 24 are the only ones that satisfy the relationship. Once you get to 25, you see that the number has a length of 26, and for all future numbers you are multiplying by more than 25, so you will be adding at least one digit every time. Therefore there are no numbers greater than 25 for which this relationship will hold.

If you would like to verify the above, you can put the numbers 1..25 in the first column of an excel spreadsheet, and then calculate how many digits the factorial of each number has with the following formula, that you can put in B1 and copy down to B25:

=CEILING(LOG(FACT(A1);10);1)

[This is mostly the same as Geoff's solution]

Let us consider the pairs $(n,n!)$ and $((n+1),(n+1)!)$ . If for some $n$ , we know that $n!$ has more digits than $n$ already, then we cannot hope that $(n+1)!$ has lesser digits than $n!$ and hence, equal to $(n+1)$ .

So, our strategy would be to search for values of $i$ such that $\text{digits}(i!) = i$ from $i=1$ onwards. Once we encounter an $i$ such that $\text{digits}(i!) > i$ , we can quit looking because of the reason above.

So, we could use the following code:

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lenDigits = lambda n: len(str(n))

def factorial(n):
    r = 1
    for i in range(1,n+1):
        r *= i
    return r

i = 1
while lenDigits(factorial(i)) <= i:
    if lenDigits(factorial(i)) == i:
        print(i)
    i += 1