Yes. From $n=(x+1)+\dots+(x+1990)=1990x+1989\cdot 995$ we get $n$ - odd and $n$ is a multiple of $995$. Suppose $n=(x_{i}+1)+\dots+(x_{i}+k_{i})$, for $i=\overline{1,1990}$ and one of the $k_{i}$'s is 1990. Also $k_{i}>1$. Then $2n=k_{i}(2x_{i}+k_{i}+1)$. For any $k_{i}|n$, $k_{i}<\sqrt{n}$ there is an integer $x_{i}=\frac{1}{2}\left(\frac{n}{d}-k_{i}-1\right)$. So $2n$ has exactly $1990$ divisors which are $>1$ and $<\sqrt{n}$, which means that $2n$ has $2\cdot1991=2\cdot11\cdot181$ divisors. Since $2n$ has already $3$ prime divisors: $2,5,199$ we conclude that these are the only prime divisors of $2n$. Now since $n$ is odd, it is clear that $n$ must be $5^{10}\cdot199^{180}$ or $5^{180}\cdot199^{10}$. So, such $n$ exists, but these are the only values $n$ can take.

yes $ 5^{10}\cdot199^{180}$ has exactly 1990 divisors, but no all are neccessery $ <\sqrt{n}% $

I didn't exactly understand your question, but I did make some typos in the solution: freemind wrote: Then $ 2n = k_{i}(2x_{i} + k_{i} + 1)$. For any $ k_{i}|n$, $ k_{i} < \sqrt {n}$ there is an integer $ x_{i} = \frac {1}{2}\left(\frac {n}{d} - k_{i} - 1\right)$. So $ 2n$ has exactly $ 1990$ divisors which are $ > 1$ and $ < \sqrt {n}$, which means that $ 2n$ has $ 2\cdot1991 = 2\cdot11\cdot181$ divisors. should be: ___ Then $ 2n = k_{i}(2x_{i} + k_{i} + 1)$. For any $ k_{i}|n$, $ k_{i} < \sqrt {2n}$ there is a positive integer $ x_{i} = \frac {1}{2}\left(\frac {2n}{k_i} - k_{i} - 1\right)$. So $ 2n$ has exactly $ 1990$ divisors which are $ > 1$ and $ < \sqrt {2n}$, which means that $ 2n$ has $ 2\cdot1991 = 2\cdot11\cdot181$ divisors. ___ Hope this helps.

Let $n$ be a number with the desired properties. By the first condition we get $n=x+(x+1)+\cdots+(x+1989)=995(2x+1989) \Rightarrow 5\mid n,199 \mid n$ and $n$ is odd. By the second condition there exists $1990$ pairs of $(j,k)$ such that $n=j+(j+1)+\cdots+(j+k)$,or $2n=(k+1)(k+2j)$.It is also easy to see that $k+1<k+2j$ so $\frac{d(2n)}{2}-1=1990 \Rightarrow d(2n)=3982$.Now it is natural to set $2n=2\times 199^{y_1} \times 5^{y_2} \times {p_1}^{y_3} \times \cdots {p_m}^{y_m}$.Then $3982=2\times 11\times 181=d(2n)=2(y_1+1)(y_2+1)\cdots(y_m+1)$.We also know that $y_1,y_2 \ge 1$.This forces us to conclude that either $y_1+1=11,y_2+1=181$ or $y_1+1=181,y_2+1=11$ and the rest $y_i=0$Thus $(y_1,y_2)=(10,180),(180,10)$ and the numbers $5^{180}\times199^{10}$ and $5^{10}\times199^{180}$ satisfy our conditions.So the answer to the problem is:Oh,Yes!

Proof (sketch): The first condition implies $n \equiv 995$ (mod 1990). So 5 and 199 appear in $n$'s prime factorization. Also note that n has to be odd. If n can be decomposed into an even number 2k, then $n \equiv k$ (mod 2k). If n can be decomposed into an odd number, d, of consecutive integers, then d $\mid$ n. We can combine these two statements to get an iff for the number of numbers in which odd n can be decomposed: All divisors of 2n work. Also note that $\sqrt{2n}$ provides the upper bound of divisors. This is convenient, since it just divides the divisors of 2n by 2. So we just need an n that prime factors 5 and 199 with 1991 divisors. This gives us the two solutions $5^{10} * 199^{180}$ and $5^{180} * 199^{10}$.

I constructed a number $x=LCM[3,5,7,....,2653]$. It probably satisfies both the conditions. Because it can be written as the sum of $1990$ consecutive integers. All can be written as the sum of $n$ consecutive integers where $n$ is any number which is not divisible by $4$ and $2\le n\le 2654$, that is exactly $1990$. Please PM me or post here if there is any mistake.