Print the number of subarrays of an array having negative sums
Consistency


                for(j=0;j&lt;i;j++){
                    sum += arr[j+k-1];
                }
                if(sum&lt;0)
                    count++;



It's more common to write the latter control structure 

                if (sum &lt; 0) count++;


That is the clearest way to write the single statement version.  

But if you're doing that, why not say 

                for (int j = 0; j &lt; i; j++) sum += arr[j+k-1];


How do you choose when to use or not use the single statement form?  

I find it easiest to never use it.  So I'd say 

                if (sum &lt; 0) {
                    count++;
                }


Then I don't have to consider whether I should use it this time or not.  

There are also advantages if someone has to edit it in the future.  In particular, it's harder to make editing mistakes.  

Keep it simple


        for(int i=1,k,j;i&lt;=N;i++){
            for(k=1;k&lt;=N-i+1;k++){
                for(j=0;j&lt;i;j++){
                    sum += arr[j+k-1];



Consider 

        for (int i = 0; i &lt; arr.length; i++) {
            for (int k = 1; k &lt;= arr.length - i; k++) {
                for (int j = 0; j &lt;= i; j++) {
                    sum += arr[j+k-1];


In the second for, you had N-i+1, which is the same as N-(i-1).  But you can change the bounds of i from 1 to N to 0 to N-1.  Then you don't need the extra 1.  But then you need to change the third for loop to j &lt;= i.  

I would rather use arr.length than N, as that's the actual limiting factor.  

I added some whitespace for readability.  

        for (int i = 0; i &lt; arr.length; i++) {
            for (int k = 0; k &lt; arr.length - i; k++) {
                for (int j = 0; j &lt;= i; j++) {
                    sum += arr[j+k];


In the summation step, you have j+k-1.  But if we change the bounds of k from 1 to N-i to 0 to N-i-1, we can just do j+k.  

I find this version easier to follow when I read it.  

        for (int i = 0; i &lt; arr.length; i++) {
            for (int k = 0; k &lt; arr.length - i; k++) {
                long sum = 0;
                for (int j = 0; j &lt;= i; j++) {
                    sum += arr[j+k];


You declare sum outside all the for loops.  That's not necessary.  You could define it just before the third.  Then you don't have to reset it at the end of that loop.  It will be automatically set whenever you need it.  

Using long instead of int avoids the possibility of overflow.  

Alternatives

If you have plenty of space or are willing to modify the input array, you can do this with only two levels of loop nesting.  You can calculate the sums prior to checking them.  

        long[][] sums = new long[arr.length][arr.length];
        for (int i = 0; i &lt; arr.length; i++) {
            sums[i][i] = arr[i];
        }

        for (int start = 0; start &lt; arr.length; start++) {
            for (int end = start + 1; end &lt; arr.length; end++) {
                sums[start][end] = sums[start][end - 1] + arr[end];
            }
        }

        int count = 0;
        for (int start = 0; start &lt; arr.length; start++) {
            for (int end = start; end &lt; arr.length; end++) {
                if (sums[start][end] &lt; 0) {
                    count++;
                }
            }
        }


But we can actually do better.  

        long[] sums = new long[arr.length];
        sums[0] = arr[0];
        for (int i = 1; i &lt; arr.length; i++) {
           sums[i] = sums[i - 1] + arr[i];
        }

        int count = 0;

        // count the arrays starting with the first element
        for (int end = 0; end &lt; arr.length; end++) {
            if (sums[end] &lt; 0) {
                count++;
            }
        }

        // count the arrays starting with start+1
        for (int start = 0; start &lt; arr.length; start++) {
            for (int end = start + 1; end &lt; arr.length; end++) {
                if (sums[end] - sums[start] &lt; 0) {
                    count++;
                }
            }
        }


Less space but the same runtime growth.  

Both solutions make use of the fact that a sum of N elements is equal to a sum of N-1 elements plus the Nth element.  So we don't have to keep recalculating that, which is the slow part of the original code.