Is $n$ times $O(1)$ equivalent to $O(n)$?
Think in terms of the definition: $f(m)=O(g(m))$ means that there exists some constants $c&gt;0, M\ge 0$ 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
$$
f_i(m) = \begin{cases}
2^i &amp; \text{if $m&lt;i$}\\
1 &amp; \text{if $m\ge i$}
\end{cases}
$$
Now each $f_i(m)\in O(1)$, since for each of them we could find a $c_i&gt;0,M_i\ge 0$ 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.