We need the sum of all integers $n$ from $1$ to $100$ such that $\gcd(n,2040)=1$.

1. Prime factorization of $2040$

$2040=204\cdot 10=(2\cdot 102)(2\cdot 5)=2^3\cdot 3\cdot 5\cdot 17$

So, for $\gcd(n,2040)=1$, the number $n$ must not be divisible by any of:

$2,\ 3,\ 5,\ 17$

Thus we need the sum of numbers from $1$ to $100$ that are not divisible by $2,3,5,$ or $17$.

2. Start with sum of first $100$ natural numbers

$1+2+\cdots+100=\frac{100\cdot 101}{2}=5050$

Now subtract the sum of numbers divisible by at least one of $2,3,5,17$ using inclusion-exclusion.

3. Sums of multiples of single primes

For multiples of $d$ up to $100$, the sum is

$d(1+2+\cdots+\lfloor 100/d\rfloor)=d\cdot \frac{k(k+1)}{2}$

where $k=\lfloor 100/d\rfloor$.

(i) Multiples of $2$

$2(1+2+\cdots+50)=2\cdot \frac{50\cdot 51}{2}=2550$

(ii) Multiples of $3$

$3(1+2+\cdots+33)=3\cdot \frac{33\cdot 34}{2}=1683$

(iii) Multiples of $5$

$5(1+2+\cdots+20)=5\cdot \frac{20\cdot 21}{2}=1050$

(iv) Multiples of $17$

$17(1+2+\cdots+5)=17\cdot \frac{5\cdot 6}{2}=255$

So,

$S_1=2550+1683+1050+255=5538$

4. Add back sums of multiples of pairwise LCMs

(i) Multiples of $\mathrm{lcm}(2,3)=6$

$6(1+\cdots+16)=6\cdot \frac{16\cdot 17}{2}=816$

(ii) Multiples of $\mathrm{lcm}(2,5)=10$

$10(1+\cdots+10)=10\cdot \frac{10\cdot 11}{2}=550$

(iii) Multiples of $\mathrm{lcm}(2,17)=34$

$34(1+2)=102$

(iv) Multiples of $\mathrm{lcm}(3,5)=15$

$15(1+\cdots+6)=15\cdot \frac{6\cdot 7}{2}=315$

(v) Multiples of $\mathrm{lcm}(3,17)=51$

$51(1)=51$

(vi) Multiples of $\mathrm{lcm}(5,17)=85$

$85(1)=85$

Thus,

$S_2=816+550+102+315+51+85=1919$

5. Subtract sums of multiples of triple LCMs

(i) Multiples of $\mathrm{lcm}(2,3,5)=30$

$30(1+2+3)=180$

(ii) Multiples of $\mathrm{lcm}(2,3,17)=102$

No multiple $\le 100$, so sum $=0$.

(iii) Multiples of $\mathrm{lcm}(2,5,17)=170$

No multiple $\le 100$, so sum $=0$.

(iv) Multiples of $\mathrm{lcm}(3,5,17)=255$

No multiple $\le 100$, so sum $=0$.

Hence,

$S_3=180$

6. Quadruple intersection

$\mathrm{lcm}(2,3,5,17)=510>100$

So the sum for quadruple intersection is $0$.

7. Inclusion-exclusion result

Sum of numbers from $1$ to $100$ divisible by at least one of $2,3,5,17$ is

$S_1-S_2+S_3=5538-1919+180=3799$

Therefore required sum is

$5050-3799=1251$

8. Final answer

$\boxed{1251}$

This matches the stored correct answer.