Think in terms of the definition: $f(m)=O(g(m))$ means that there exists some constants $c>0, M\ge 0$ such that $f(m) \le c\cdot g(m)$ for all $m\ge M$ so when you write $$ \sum_{i=1}^nO(1) $$ you mean that there are $n$ functions, $f_1, f_2, \dotsc, f_n$ and constants $c_1, c_2, \dotsc, c_n$ such that $f_i(m)\le c_i\cdot1$ for all sufficiently large values of $m$. Your sum then is $$ \sum_{i=1}^nO(1)=\sum_{i=1}^nf_i(m)\le \sum_{i=1}^nc_i\le \max_{1\le j\le n}\{c_j\}\cdot n = Kn $$ for all sufficiently large $m$. But this is, by definition, the same as saying $$ \sum_{i=1}^nO(1) = O(n) $$

Think in terms of the definition: $f(m)=O(g(m))$ means that there exists some constants $c>0, M\ge 0$ such that $f(m) \le c\cdot g(m)$ for all $m\ge M$ so when you write $$ \sum_{i=1}^nO(1) $$ you mean that there are $n$ functions, $f_1, f_2, \dotsc, f_n$ and constants $c_1, c_2, \dotsc, c_n$ such that $f_i(m)\le c_i\cdot1$ for all sufficiently large values of $m$. Your sum then is $$ \sum_{i=1}^nO(1)=\sum_{i=1}^nf_i(m)\le \sum_{i=1}^nc_i\le \max_{1\le j\le n}\{c_j\}\cdot n $$ for all sufficiently large $m$. The problem here is, as Hendrik and babou pointed out, the $\max$ term is a function of $n$. For instance, if $f_i(m) = i$ were constant functions, individually they are all in $O(1)$ but their sum $$ F(n) = \sum_{i=1}^nf_i(m)= \sum_{i=1}^ni = \frac{n(n+1)}{2}\notin O(n) $$ The constants are not the only problem here, though, even if we restricted them all to be bounded by some constant $K$. Suppose we had defined the functions $f_i(m)$ to be $$ f_i(m) = \begin{cases} 2^i & \text{if $m<i$}\\ 1 & \text{if $m\ge i$} \end{cases} $$ Now each $f_i(m)\in O(1)$, since for each of them we could find a $c_i>0,M_i\ge 0$ such that for all $m\ge M_i$ we have $f_i(m)\le c\cdot 1$ (just pick $c_i=1$ and $M=i$ for instance).

In this case, we also don't have $F(n)\in O(n)$: even though all the $c_i = 1$ we can't now find a fixed N so that $F(n)\le Kn$ for all $n\ge N$, since we chose the functions $f_i$ so that the "crossing point", $M_i$, is a moving target. No matter what $N$ you pick, there will be a point where beyond $N$ the sum will be larger than $Kn$.

It seems that the best you can say is that $F(n)=O(n)$ when all of the $c_i$ are less than or equal to a single fixed bound and the same holds for all the $M_i$. Useful in some cases, I suppose, but woefully wrong in general.

Think in terms of the definition: $f(n)=O(g(n))$ means that there exists some constants $c>0, N\ge 0$ such that $f(n) \le cg(n)$ for all $n\ge N$ so when you write $$ \sum_{i=1}^nO(1) $$ you mean that there are $n$ functions, $f_1, f_2, \dotsc, f_n$ and constants $c_1, c_2, \dotsc, c_n$ such that $f_i(n)\le c_i\cdot1$ for all sufficiently large values of $n$. Your sum then is $$ \sum_{i=1}^nO(1)=\sum_{i=1}^nf_i(n)\le \sum_{i=1}^nc_i\le \max_{1\le j\le n}\{c_j\}\cdot n = Kn $$ for all sufficiently large $n$. But this is, by definition, the same as saying $$ \sum_{i=1}^nO(1) = O(n) $$

Think in terms of the definition: $f(m)=O(g(m))$ means that there exists some constants $c>0, M\ge 0$ such that $f(m) \le c\cdot g(m)$ for all $m\ge M$ so when you write $$ \sum_{i=1}^nO(1) $$ you mean that there are $n$ functions, $f_1, f_2, \dotsc, f_n$ and constants $c_1, c_2, \dotsc, c_n$ such that $f_i(m)\le c_i\cdot1$ for all sufficiently large values of $m$. Your sum then is $$ \sum_{i=1}^nO(1)=\sum_{i=1}^nf_i(m)\le \sum_{i=1}^nc_i\le \max_{1\le j\le n}\{c_j\}\cdot n = Kn $$ for all sufficiently large $m$. But this is, by definition, the same as saying $$ \sum_{i=1}^nO(1) = O(n) $$