Arithmetic growth versus exponential decay

Looking at this with no equations, a simple thought experiment. If, in fact, it grew without bounds, at some point, you have over 800kg. And with a 1 yr half-life, it will drop to 400kg a year later. Now, during that year, you've only added 365kg, and it's been there an average of 6 months so .707 remains (the square root of 'half'), say 30% gone or about 250 remaining. We now have just 650kg.

The other answers clearly show precise numbers, but when I read your question, it seemed suited to a simple back of envelope approach to simply prove the negative.

Suppose that at a certain stage we have an amount $M$ of the substance in the container. Then after $1$ day, the amount left is $M(1/2)^{1/365}$, so the amount that has died is $M(1-(1/2)^{1/365})$. If $M$ is large enough, the amount that has died in $1$ day is greater than $1$ kilogram. So the added $1$ kilogram cannot compensate for the amount that has died.

It follows that the amount of the substance in the container can never be greater than $M$. To be explicit, the amount cannot ever be greater than $\dfrac{1}{1-(1/2)^{1/365}}$.

Let's model this as a recursive sequence $X_1, X_2, \cdots, X_n, \cdots$, where $X_n$ is the number of kilograms of the element in the contains on the $n$th day. Since the half life of the element is $1$ year, after a single day a proportion of $$\sqrt[365]{\frac{1}{2}}\approx 0.998$$ of the substance you began with will remain. Since we also add $1$ kilogram after each day, the amount of the elment on the $n+1$st day will be $$X_{n+1} = \sqrt[365]{\frac{1}{2}}X_n + 1$$ $$= \sqrt[365]{\frac{1}{2}}\left(\sqrt[365]{\frac{1}{2}}X_{n-1} + 1\right) + 1 = \frac{1}{2^{2/365}}X_{n-1} + \frac{1}{2^{1/365}} + 1$$ $$\vdots$$ $$= \frac{1}{2^{(n+1)/365}} + \frac{1}{2^{n/365}} + \cdots + \frac{1}{2^{1/365}} + 1$$ This is a geometric sum, which we can express as $$X_{n+1} = \frac{1-2^{-(n+2)/365}}{1-2^{1/365}}.$$ So, substituting $n$ for $n+1$, $$X_n = \frac{1-2^{-(n+1)/365}}{1-2^{1/365}}.$$ Taking the limit as $n$ becomes very large, we find $$\lim_{n\to \infty} X_n = \frac{1}{1-2^{1/365}} \approx 527,$$ so, after the process goes on for a very long time, the amount of radioactive material you have will be bounded, never growing to be much larger than $527$ (i.e., the exponential decay wins out over the arithmetic growth).