It's as easy as 1, 3, 5

$10=2 (mod 4), 13 = 1 ( mod 4)$ we get 1+2=3, if n is an even, then $10n^2 = 0(mod 4)$ why, $10= 5 \times 2$ an even has 2 as a divisor, so $10n^2= 5 \times 2 \times 2 \times x$ $10n^2 =5\times 4x, n=0 (mod 2)$ it is divide able by 4, and would become 1 (m0d 4) so it is $n=1 (Mod 2)$ when an odd no. is divided by 4 the results are 1,3. so 1+3 $\boxed{4}$

As upon division of $10{ n }^{ 2 }+13$ by 4 has the remainder 3, We can say that, $10{ n }^{ 2 }+13-3$ is divisible by 4 ! Hence, $4|10{ n }^{ 2 }+10$ or, $4|8({ n }^{ 2 }+1)+2({ n }^{ 2 }+1)$

Now , As the 5th rule of Divisibility, which is
If x|y and x|y+z then, x|z . By this rule we can say that, $4|2({ n }^{ 2 }+1)$ hence, $2|({ n }^{ 2 }+1)$ So, n^2+1 must be an even number . Here ${ n }^{ 2 }+1$ will be even only if n^2 is odd So, the possible values of n are {1,3,5,7,9.......} . *As there is no even value of n , the only possible remainders of n upon division by 4 are 1 and 3 *

Hence , The Sum of the possible remainders is 1+3=4

Since $\quad 10n^2+13 \equiv 3 \pmod {4} \quad \Rightarrow 4|10n^2+10 \quad \Rightarrow 4|2n^2+2$ .

For $2n^2+2$ to be divisible by $4$ , $n$ must be odd. As $n$ is odd then the possible remainders when divided by $4$ are $1$ and $3$ , therefore the answer is $1+3 = \boxed {4}$ .