Suppose $m=2$. If $k$ is odd, and $n$ is even, then there are involutions switching even numbers and odd numbers up to $n$ such as $1 \leftrightarrow 2, 3 \leftrightarrow 4 ...$. Letting one of these act on subsets of size $k$ switches the parity of the sum, so there are equally many subsets of size $k$ with an odd sum as with an even sum. For no other cases with $m=2$ are the subsets divided equally by the parities of their sums.

Let $f_2(n,k) = \# \text{subsets of } \lbrace 1, ..., n\rbrace ~\text{of size } k ~\text{with even sum}$.

Let $g_2(n,k) = 2f_2(n,k) - {n\choose k},$ the count with even sum minus the count with odd sum.

$g_2(2a, 2c) = (-1)^c {a \choose c}$

$g_2(2a, 2c+1) = 0$

$g_2(2a+1,2c) = g_2(2a,2c)$

$g_2(2a+1,2c+1) = -g_2(2a,2c)$

For example, let $n=30$ and $k=10$. There are $30,045,015$ subsets of $\lbrace 1, 2, ..., 30\rbrace$ size $10$, of which $f_2(30,10) = 15,021,006$ have an even sum and $15,024,009$ have an odd sum. The difference is $-{15 \choose 5} = -3,003$.

To prove these formulas, consider the group action of $C_2^a$ with generators switching $1 \leftrightarrow 2, ..., 2a-1\leftrightarrow 2a$. If for some pair $(2t-1,2t)$ only one of these is in a subset, then the orbit of that subset has equally many subsets with even and odd sums since switching $2t-1 \leftrightarrow 2t$ changes the parity of the sum. What is left over is the subsets which contain both $2t-1$ and $2t$ or neither, and there are $a \choose c$ of these left over. The parities of these are all the same, and the parity is determined by the parities of $c, n, k$ as indicated above.

Suppose $p$ is an odd prime.

Let $f_p(n,k)$ be the number of subsets of $\lbrace1,...,n\rbrace$ of size $k$ with sum divisible by $p$. If $n$ is divisible by $p$, and $k$ is not, then $f_p(n,k) = {n\choose k}/p$, since applying the permutation $(1 ~2~ ...~ p)(p+1 ~~p+2~~ ...~ 2p)...$ to a subset of size $k$ adds $k$ to the sum $\mod p$, and repeating this hits every congruence class once in each orbit.

Let $g_p(n,k) = p f_p(n,k) - {n\choose k}.$

Let $n=p a + b, k = pc + d$, with $0 \le b,d \lt p$. Consider the action of $C_p^a$ generated by $(1 2 ... p), (p+1 ~~ p+2 ~ ... ~2p),...$. If the intersection of a subset with $\lbrace (t-1)p+1, (t-1)p+2, ... tp \rbrace$ has size from $1$ to $p-1$, then its orbit is evenly split among the congruence classes. So, the imbalance comes from the subsets containing $c$ complete blocks of size $p$, together with some subset of size $d$ of the last $b$ elements. Note that if $d \gt b$ then this is impossible so $g_p(n,k) = 0$.

$g_p(pa+b,pc+d) = {a \choose c} g_p(b,d)$.

Crude estimates for $g_p(n,k)$ with $n,k \lt p$, such as $|g_p(n,k)| \le p 2^n$ turn into general bounds for larger $n,k$. $$|g_p(pa+b,pc+d)| \le {a \choose c} p 2^b.$$

One simplification compared with $m=2$ is that $1+2+...+p$ is divisible by $p$ when $p$ is an odd prime. That $1+2$ is odd produced the $(-1)^c$ factor.

If $m$ is composite, then some of the arguments used for $m$ prime don't work.