Its the new millennium! Oh wait... that's a bit late.

So we know we have a series $\underbrace{m+m+1+m+2+m+3+\ldots}_\text{n times}=2000.$ We use the arithmetic progression formula $S_n=\frac{n}{2}(2a+(n-1)d)$ for this series. So we get $\frac{n}{2}(2m+n-1)=2000$ because d=1 and a=m. By rearranging the equation, we get $2mn+n^2-n=4000$ which is rearranged and factorised to $n^2+(2m-1)n-4000=0.$ We notice that 2m-1 is trivially odd. So we want to factorise this equation to the form of $(n ? ? ? )(n ? ? ? )$ where we need to fill in the ???s. We actually have many solutions for this, but we need the case in which we have the smallest m. So, to minimise m, we first factorise 4000 to get $4000=2^5\times5^3.$ Now we realise there are only a few combinations, namely $5\times800, 25\times160 \text{ and} 125\times32.$ Then, we realise we need the pair with the least difference so we can minimise m. So we have the factorisation $(n-32)(n+125)=0.$ So we have $n^2+93n-4000=0$ and thus leading to $2m-1=94$ and thus $m=\boxed{47}$

Same logic as Joshua's solution but perhaps better explanation on how to get the final solution.

From: $m + (m+1) + (m+2) + ... + (m+n-1) = 2000$

$\Rightarrow \dfrac {n(2m+n-1)} {2} = 2000$

$\Rightarrow 2m = \dfrac {4000}{n} - n + 1$

Our task is to find the largest $n$ that satisfies the above equation.

We note that since $LHS$ is always even, $RHS$ must also be even. We also note that $4000 = 2^5 \times 5^3$ . When $n$ is odd, that is $n = 5, 25, 125$ , then both $\frac{4000}{n}$ and $n+1$ are even. We notice that $n = 125$ gives a negative $m$ which is unacceptable. When $n=25$ , $m = 68$ .

Now, consider when $n$ is even. Then $n+1$ is odd and $\frac {4000} {n}$ has to be odd too. This will only happen when $n=32$ and then $\frac{4000}{n} = 125$ which is odd. Therefore,

$\Rightarrow 2m = \dfrac {4000}{32} - 32 + 1 = 125 - 32 + 1 = 94$

$\Rightarrow m = \boxed{47}$ , which is smaller than $68$ .