Ho to write the negation of the statement? I would like to prove the following fact using contradiction.
${\rm C{\small LAIM}}$: Let $\mathbf{M}=[m_{ij}]$ be a $n\times n$ matrix of non-negative entries. Assume $\sum_{i=1}^{n}\sum_{j=1}^{n}m_{ij}\leqslant \gamma n$ for some $\gamma\geqslant0$. Then, for any $\lambda>1$, there are at least $(1-\frac{1}{\lambda})n$ rows (columns) $\mathbf{r}$ of $\mathbf{M}$ for which $\sum_{i=1}^{n}r_{i}\leqslant \gamma\lambda$.
I would like to assume the opposite. So, my question is what is the negation of 

for any $\lambda>1$, there are at least $(1-\frac{1}{\lambda})n$ rows (columns) $\mathbf{r}$ of $\mathbf{M}$ for which $\sum_{i=1}^{n}r_{i}\leqslant \gamma\lambda$?

Is it 

for any $\lambda>1$, there are at most $(1-\frac{1}{\lambda})n$ rows (columns) $\mathbf{r}$ of $\mathbf{M}$ for which $\sum_{i=1}^{n}r_{i}> \gamma\lambda$?

 A: The negation of

for any $\lambda>1$, there are at least $(1-\frac{1}{\lambda})n$ rows (columns) $\mathbf{r}$ of $\mathbf{M}$ for which $\sum_{i=1}^{n}r_{i}\leqslant \gamma\lambda$

is:

There exists $\lambda>1$, such that the number $N$ of rows (columns) $\mathbf{r}$ of $\mathbf{M}$ for which $\sum_{i=1}^{n}r_{i}\leqslant \gamma\lambda$ satisfies $N < (1-\frac{1}{\lambda})n$.

You have to be a bit careful about the negation of "at least". I supposed it means $N \geqslant (1-\frac{1}{\lambda})n$, but if you mean $>$, the condition on $N$ in the negation should be $N \leqslant (1-\frac{1}{\lambda})n$.
A: Anytime I want to negate a statement like that I first translate it into pure symbols. Here is my attempt at translation.

for any $\lambda>1$, there are at least $(1-\frac{1}{\lambda})n$ rows (columns) $\mathbf{r}$ of $\mathbf{M}$ for which $\sum_{i=1}^{n}r_{i}\leqslant \gamma\lambda$

\begin{align}
\forall\lambda\Big[ \lambda>1\implies \big|\{\mathbf{r}\in\text{Rows}(\mathbf{M}): \sum_{i=1}^{n}r_{i}\leqslant \gamma\lambda\}\big| \ge (1-\frac{1}{\lambda})n \Big]
\end{align}
Lets simplify this a bit and let $S=\{\mathbf{r}\in\text{Rows}(\mathbf{M}): \sum_{i=1}^{n}r_{i}\leqslant \gamma\lambda\}$ and $k=(1-\frac{1}{\lambda})n$.
So, assuming my translation into symbols is correct, the negation is
\begin{align}
\lnot&\forall\lambda\Big[ \lambda>1\implies |S| \ge k \Big] \\
%
&\exists\lambda\Big[ \lnot\big(\lambda>1\implies |S| \ge k\big) \Big]\\
%
&\exists\lambda\Big[ \big(\lambda>1\big) \land \lnot\big(|S| \ge k\big) \Big]\\
%
&\exists\lambda\Big[ \big(\lambda>1\big) \land \big(|S| < k\big) \Big].
\end{align}
And so we see we can translate this back into English as

There exists a $\lambda>1$ such that the number of elements in $S$ is less than $k$.

Or, going further,

There exists a $\lambda>1$ such that the number of rows $\mathbf{r}$ of $\mathbf{M}$ such that $\sum_{i=1}^{n}r_{i}\leqslant \gamma\lambda$ is less than $(1-\frac{1}{\lambda})n$.

One thing to notice is that the negation didn't effect anything going on in the set $S$. This can potentially be trouble spot if trying to negate the original English statement directly.
