Let $$S_n=\sum_{k=0}^{n-1}\left\lfloor \frac kc\right\rfloor$$
and
$$ T_n=\sum_{k=0}^{n-1}S_k.$$
Then we are looking for
$$ \begin{align}\sum_{k=0}^N\sum_{l=0}^M\left\lfloor\frac{k+l}{c}\right\rfloor
&=\sum_{k=0}^N(S_{M+k+1}-S_k)\\
&=T_{N+M+2}-T_{M+1}-T_{N+1}\end{align}$$
Note that for $n=qc+r$ with $0\le r<c$, we have
$$S_n=c\sum_{k=0}^{q-1}k+rq=\frac c2{q(q-1)}+rq.$$
By the same trick,
$$\begin{align}T_n&=\sum_{k=0}^{q-1}\left(\frac c2{k(k-1)}+\frac{k(k-1)}2k\right)+r\cdot\frac c2{q(q-1)} +\frac{r(r-1)}2q\\
&=\frac {cq(q-1)(q-2)}6+\frac{q(q-1)(q-2)(3q-1)}{12}+\frac{rq(cq-c+r-1)}2\end{align}$$
If you sub $u=1-t$, you have
$$\int_{u=1}^{u=0}(-1)^{\lfloor{1994-1994u\rfloor}} (-1)^{\lfloor{1995-1995u\rfloor}} \binom{1993}{\lfloor{1994-1994u\rfloor}}\binom{1994}{\lfloor{1995-1995u\rfloor}} (-du)$$
Now note that there are values of $u$ where $\lfloor{1994-1994u\rfloor}$ is an integer, and at such places $\lfloor{1994-1994u\rfloor}\neq1993-\lfloor{1994u\rfloor}$. However those places have measure $0$ and contribute nothing to the integral. Everywhere else, where $\lfloor{1994-1994u\rfloor}$ is not an integer, then $\lfloor{1994-1994u\rfloor}=1993-\lfloor{1994u\rfloor}$.
And there is the analogous statement for $1995$. So you have:
$$-\int_{u=1}^{u=0}(-1)^{1993-\lfloor{1994u\rfloor}} (-1)^{1994-\lfloor{1995u\rfloor}} \binom{1993}{1993-\lfloor{1994u\rfloor}}\binom{1994}{1994-\lfloor{1995u\rfloor}} du$$
And for more basic reasons, this is equal to
$$-\int_{u=0}^{u=1}(-1)^{\lfloor{1994u\rfloor}} (-1)^{\lfloor{1995u\rfloor}} \binom{1993}{\lfloor{1994u\rfloor}}\binom{1994}{\lfloor{1995u\rfloor}} du$$
This is the negative of what we started with, using $u$ as the integration variable instead of $t$. So it all must be $0$.
Best Answer
That's an excellent suggestion! Ceiling and floor functions are awkward, so my first step is always to remove them somehow:
Even $p$
For even $p$, we can define $p=2m$. Then the suggestion given to you become:
$$ \begin{aligned} \frac{1}{2}\left( \sum_{k=0}^{m}\binom{2m}{k} + \sum_{k=0}^{m}\binom{2m}{2m-k} \right) &= \frac{1}{2}\left( \sum_{k=0}^{m}\binom{2m}{k} + \sum_{k=m}^{2m}\binom{2m}{k} \right) \\\\ &= \frac{1}{2}\left( \binom{2m}{m}+\sum_{k=0}^{2m}\binom{2m}{k} \right) \\\\ &= \frac{1}{2}\left( \binom{p}{\frac{p}{2}}+2^{p} \right) \end{aligned} $$
Odd $p$
We define $p=2m+1$. Then the suggestion becomes:
$$ \begin{aligned} \frac{1}{2}\left( \sum_{k=0}^{m}\binom{2m+1}{k} + \sum_{k=0}^{m}\binom{2m+1}{2m+1-k} \right) &= \frac{1}{2}\left( \sum_{k=0}^{m}\binom{2m+1}{k} + \sum_{k=m+1}^{2m+1}\binom{2m+1}{k} \right) \\\\ &= \frac{1}{2}\left( \sum_{k=0}^{2m+1}\binom{2m+1}{k} \right) \\\\ &= \frac{1}{2}\cdot 2^{p} \end{aligned} $$
Key Points